Are the Milnor's seven dimensional exotic spheres parallelizable?
A much more general result is true.

Theorem: Let $\Sigma$ be a homotopy sphere and $f: S^n \to \Sigma $ be a homotopy equivalence. Then $f^{\ast} T \Sigma \cong T S^n$.

It says that exotic spheres cannot be distinguished by looking at the tangent bundle.
This result is one of the hidden gems of the golden age of topology and the proof invokes the whole plethora of topology of the 1950s.

The argument can be recollected from the old literature, but I do not know a coherent reference.

To start with, there are several invariants of the tangent bundles that do not depend on the smooth structure. 
Let $\Sigma$ be a homotopy sphere. Then:


the Euler class $\chi(T\Sigma^{n})$ is $2$ if $n$ is even (Gauss-Bonnet, relatively easy).
$T \Sigma^n \oplus  \mathbb{R}$ is trivial. This is a deep result by Kervaire and Milnor (not in the Annals paper, but a small note published before). $T \Sigma \oplus \mathbb{R}$ is given by an element of $\pi_{n-1} (O)$, which is known, by Bott periodicity, to be either $Z$, $0$ or $Z/2$. The $Z$ groups are detected by the Pontrjagin class, which has to vanish by Hirzebruchs signature formula because the sphere evidently has signature $0$. In the $Z/2$ case, the argument is more delicate. Essentially, the normal spherical fibration of a manifold does not depend on the smooth structure. Since the normal fibration of the standard sphere is trivial, so is that of $T \Sigma$. The process of associating to a vector bundle its spherical fibration is the J-homomorphism which is injective in these dimensions by Adams' $J(X)$ paper.


Now look at the homotopy sequence of the fibration $O(n)\to O(n+1) \to S^n$, i.e. the piece

$$
\mathbb{Z}=\pi_n (S^n) \to \pi_{n-1} (O(n)) \to \pi_{n-1} (O(n+1)) = \pi_{n-1} (O). 
$$

It is known that $TS^n$ (for the standard smooth structure) is the image of the generator of $\pi_n (S^n)$ (not hard, see Steenrods book).
By the above deep result, $T \Sigma$ lies in the kernel of
$\pi_{n-1}(O(n)) \to \pi_{n-1} (O(n+1))$, i.e. it also comes from $Z=\pi_n (S^n) $. The image of $Z \to \pi_{n-1} (O(n)$ can be computed.
It is $Z$ if $k$ is even (using the Euler class),
it is $0$ if $n=1,3,7$ (follows directly from Adams' result on the parallelizability of the standard spheres) and it is $Z/2$ in the 
remaining cases.
You can find the (not so hard, but clever) argument for the last assertion in Levine's lectures on homotopy spheres (which can be viewed as the sequel to the Kervaire-Milnor paper).

How to proceed? If $n=1,3,7$, it follows that $T \Sigma$ is trivial, as $TS^n$. If $n$ is even, then the kernel of 
$\pi_{n-1}(O(n))\to \pi_{n-1}(S^{n+1})$ is detected by the Euler class,
and by Gauss-Bonnet, this characterizes $T \Sigma$.

It remains the case of odd $n$ apart from the "Adams dimensions". One has to argue that in these dimensions, $T \Sigma$ is nontrivial.
In the introduction to his Hopf invariant paper, Adams attributes to Dold the result ''$T \Sigma$ parallel implies that $\Sigma$ (and hence $S^n$) is an H-space''.
But he (Adams) proved that is not the case $n=1,3,7$. Adams does not give a reference for Dolds result, but in his answer to this question (and the subsequent comments), John Klein sketches a proof that looks like a 1950s argument.

EDIT: I supervised a Master's thesis (written by Julia Heller), who worked out the details of this argument.
Here is the argument for the fact that if a homotopy sphere $\Sigma^n$ is parallelizable, then $n=0,1,3,7$. 

Consider the diagonal $\Sigma \subset \Sigma \times \Sigma$. Its normal bundle $N$ is isomorphic to $T \Sigma$, 

hence trivial. We have the Pontrjagin-Thom collapse
$c:\Sigma \times \Sigma \to Th (N) $. Any trivialization $N \cong \Sigma \times \mathbb{R}^n$ induces a map $a: Th (N) \to S^n$. Let
$$
\mu:  \Sigma \times \Sigma \to Th (N) \to S^n
$$
be the composition of the maps just explained. By counting intersections, it can be seen that the restriction of $\mu$ to the submanifolds $\{x\} \times \Sigma$ and $\Sigma \times \{x\}$ has degree $\pm 1$, hence is a homotopy equivalence. By composing with $(h,k):S^n \times S^n \to \Sigma \times \Sigma$ for suitably chosen homotopy equivalences, we get a map 
$$
\mu' : S^n \times S^n \to S^n
$$
which restricts to degree 1 maps $S^n \times \{x\} \to S^n$ and $\{x\} \times S^n \to S^n$. Since maps of degree 1 are homotopic, it follows that $\mu'$ gives $S^n $ the structure of an $H$-space. By Adams' theorem, $n=0,1,3,7$. 

Note: it is not important for Adams' theorem that the $H$-space structure is homotopy associative, but homotopy unitality is essential.
