The question briefly: 

Can one explain the "Dzhanibekov effect" (see youtube videos from space station or comments below) on the basis of the standard rigid body dynamics using Euler's equations? (Or explain that this is impossible and that would be yet another focus ... )

Here are more details.

See these curious videos from a space station: 

https://www.youtube.com/watch?v=L2o9eBl_Gzw

Or in the original form which is more striking (~30 seconds after the start):
https://www.youtube.com/watch?v=dL6Pt1O_gSE

A mathematical description of the model is the standard rigid body motion in empty space (since we are in outer space), which (as is well-known) can be decomposed into a center of mass motion and a rotation part described by Euler's equations. (Euler's equations can be seen as a geodesic flow for some left-invariant metric on SO(3) - as V.I. Arnold taught.)
Let us forget about the center of mass motion.

The phenomena shown in the videos is the following -  a rigid body is rotating around the axis and then SUDDENLY the rotation axis CHANGES ITS POSITION by 180 degrees. And this happens periodically with some time period.

I am a little puzzled how to explain this.

If we admit that the motion of this rigid body is EXACTLY the rotation around the axis this is for sure IMPOSSIBLE - Euler's equations predict that such a motion will continue forever. However in the real world there is nothing exact so it might be related to some instability effect - that we are around an unstable equilibrium - and go away from it after some time...

Can it be so?

If it so, it is however not clear why it happens so quickly and we get exactly the rotation axis changing its position by 180 degrees...



Vladimir Dzhanibekov who observed this in 1985 is a famous Russian cosmonaut 
https://en.wikipedia.org/wiki/Vladimir_Dzhanibekov
He was in space 5 times (as far as I know he is the champion in this).
In 1985 his  "mission impossible" was saving the Soviet space station "Salut-7" which due to some problems had run out of control...



If this can be explained I think it can be a beautiful illustration for students in lectures 
on rigid body mechanics... Also if one would want to add some humor in such a lecture one may add that our Earth is this kind of finger nut so it might also do such things ...
(Maybe it really can?)

One can see this effect qualitatively from Newtonian first principles such as $F=ma$ (as opposed to Hamiltonian or Lagrangian principles, such as conservation of energy and angular momentum) by looking at a degenerate case, when one moment of inertia is very small and the other two are very close to each other.

More specifically, consider a thin rigid unit disk, initially oriented in the $xy$ plane and centred at the origin $(0,0,0)$.  We make the "spherical cow" hypotheses that this disk has infinitesimal thickness and mass, but infinite rigidity.  On this disk, we place heavy point masses of equal mass $M$ at the points $(1,0,0)$ and $(-1,0,0)$ on the $x$ axis, and light point masses of equal mass $m$ at the points $(0,1,0)$ and $(0,-1,0)$ on the $y$ axis.  Here $0 < m \ll M$, i.e. $m$ should be viewed as negligible with respect to $M$.  (The moments of inertia are then $2m, 2M, 2(m+M)$, though we will not explicitly use these moments in the analysis below.)

We now set up the unstable equilibrium by rotating the disk around the $y$ axis.  Thus, the light $m$-masses stay fixed on the $y$-axis, while the heavy $M$-masses rotate in the $xz$-plane.  This is in equilibrium: there are no net forces on the $m$-masses, while the rigid disk exerts a centripetal force on the $M$-masses that keeps them moving in a circular motion on the $xz$-plane.

We can view this equilibrium in rotating coordinates, matching the motion of the $M$-masses.  (Imagine a camera viewing the disk, rotating around the $y$-axis at exactly the same rate as the disk is rotating.)  In this rotating frame, the disk is now stationary (so the $m$-masses are stuck on the $y$-axis at $(0,\pm 1,0)$ and the $M$-masses are stuck on the $x$-axis at $(\pm 1,0,0)$), but there is a centrifugal force exerted on all bodies proportional to the distance to the $y$-axis.  The $m$-masses are on the $y$-axis and thus experience no centrifugal force; but the $M$-masses are away from the $y$-axis and thus experience a centrifugal force, which is then balanced out by the centripetal forces of the rigid disk.

Now let us perturb the disk a bit, so that the $m$-masses and $M$-masses are knocked a little bit out of position (but keeping the centre of mass fixed at $(0,0,0)$).  In particular, the $m$-masses are knocked away from the $y$-axis and now experience a little bit of centrifugal force.  On the other hand, the rigid disk forces the light $m$-masses to remain orthogonal to the heavy $M$-masses, by exerting tension forces between the masses.  In the regime where $m$ is negligible compared to $M$, these tension forces will barely budge the heavy $M$ masses (which therefore remain essentially fixed at $(\pm 1,0,0)$ in the rotating frame), so the effect of these tension forces is to constrain the $m$-masses to lie in the $yz$-plane (up to negligible errors which we now ignore).  Rigidity also keeps the $m$-masses at a unit distance from the origin, and antipodal to each other, so the $m$-masses are now constrained to be antipodal points on the unit circle in the $yz$-plane.  However, other than this, rigidity imposes no further constraints on the location of the $m$-masses, which can then move freely as antipodal points in this unit circle.

The effect of centrifugal force in the rotating frame is now clear: if an $m$-mass (and its antipode) is perturbed to be a little bit off the $y$-axis in this unit circle with no initial velocity, then centrifugal force will nudge it a little further off the $y$-axis, slowly at first but with inexorable acceleration.  Eventually it will shoot across the unit circle and then approach the antipode of its previous position.  At this point the centrifugal forces act to slow the $m$-masses down, reversing all the previous acceleration, until one ends up with no velocity at a small distance from the antipode.  The process then repeats itself (imagine a marble rolling frictionlessly between two equally tall hills, starting from a position very close to the peak of one of the hills).

UPDATE, September 2019: Due to renewed interest in this question, I will expand upon my 2014 comment regarding why the above analysis appears at first glance to also lead to the incorrect conclusion that the disk rotation is also unstable if it one instead rotates around the $x$-axis (so that it is now the $M$-masses that are stationary and the $m$-masses that are rotating), or equivalently if one swaps the location of the $m$-masses and $M$-masses (which we will not do here to try to reduce confusion).

The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that are introduced when one is in a steadily rotating frame.  The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$, which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation).  Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame.  As it turns out, the Coriolis force has a negligible impact on the dynamics when rotating around the $y$-axis, but dominates the dynamics when rotating around the $x$-axis, which is why the preceding discussion is accurate in the former case but not the latter.  

In more detail: suppose we are rotating around the $y$-axis as in the above discussion.  In the rotating frame of reference, and starting with a configuration slightly out of equilibrium, we have as before that the $m$-masses experience a little bit of centrifugal force and begin to slide away from the $y$-axis and into the rest of the $yz$-plane.  When doing so, they will then experience some Coriolis force in a direction parallel to the $x$-axis (which direction it is depends on the orientation of the rotation, as per the right hand rule formula for the cross product).  However, due to the rigidity of the disk, as mediated by tension forces within the disk, it is not possible for the $m$-masses to actually move in the $x$-direction without also moving the much heavier $M$-masses.  But the magnitude of the Coriolis force is proportional to the small mass $m$ rather than the large mass $M$, so by Newton's law $F=Ma$ for the $M$-masses, the Coriolis force (or more precisely, the tension force produced in response to the Coriolis force) actually barely affects the motion of the $M$-masses, which basically stay put on the $x$-axis, and the $m$-masses therefore remain essentially constrained to the $yz$-plane and cannot actually experience any significant motion in the direction of the Coriolis force.  (This is what the sentence in the original explanation regarding how tension forces "barely budge" the $M$-masses was referring too, albeit somewhat obliquely.)  The analysis of the original explanation now proceeds as before.

In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$-axis.  Working in a rotating frame around this axis, the $M$-masses are near the $x$-axis of rotation, while the $m$-masses lie near the $y$-axis.  As before, the $M$-masses experience centrifugal force and thus begin drifting slightly away from the $x$-axis into the $xz$-plane.  But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$.  As such, the rigid tension forces connecting the $M$-masses to the much lighter $m$-masses offer very little resistance to the Coriolis force, and the motion of the $M$-mass begins to rotate out of the $xz$-plane, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion.  This effect tends to make the $M$-mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$-axis back into inward motion.  The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling frictionlessly around the trough of a valley).
