For example, the square can be described with the equation $|x| + |y| = 1$. So is there a general equation that can describe a regular polygon (in the 2D Cartesian plane?), given the number of sides required?

Using the Wolfram Alpha site, this input gave an almost-square:
PolarPlot(0.75 + ArcSin(Sin(2x+Pi/2))/(Sin(2x+Pi/2)*(Pi/4))) (x from 0 to 2Pi)

This input gave an almost-octagon:
PolarPlot(0.75 + ArcSin(Sin(4x+Pi/2))/(Sin(4x+Pi/2)*Pi^2)) (x from 0 to 2Pi)

The idea is that as the number of sides in a regular polygon goes to infinity, the regular polygon approaches a circle. Since a circle can be described by an equation, can a regular polygon be described by one too? For our purposes, this is a regular convex polygon (triangle, square, pentagon, hexagon and so on).

It can be assumed that the centre of the regular polygon is at the origin $(0,0)$, and the radius is $1$ unit.

If there's no such equation, can the non-existence be proven? If there are equations, but only for certain polygons (for example, only for $n < 7$ or something), can those equations be provided?

Any polygon (regular or not) can be described by an equation involving only absolute values and polynomials. Here is a small explanation of how to do that.

Let's say that a curve $C$ is given by the equation $f$ if we have $C = \{(x,y) \in \mathbb{R}^2, \, f(x,y) = 0\}$.


If $C_1$ and $C_2$ are given by $f_1$ and $f_2$ respectively, then $C_1 \cup C_2$ is given by $f_1 . f_2$ and $C_1 \cap C_2$ is given by $f_1^2 + f_2^2$ (or $|f_1| + |f_2|$). So if $C_1$ and $C_2$ can be described by an equation involving absolute values and polynomials, then so do $C_1 \cup C_2$ and $C_1 \cap C_2$.
If $C = \{(x,y) \in \mathbb{R}^2, \, f(x,y) \ge 0\}$, then $C$ is given by the equation $|f|-f$.


Now, any segment $S$ can be described as $S = \{(x,y) \in \mathbb{R}^2, \, a x + b y = c, \, x_0 \le x \le x_1, \, y_0 \le y \le y_1\}$, which is given by a single equation by the above principles. And since union of segments also are given by an equation, you get the result.

EDIT : For the specific case of the octagon of radius $r$, if you denote $s = \sin(\pi/8)$, $c = \cos(\pi/8)$, then one segment is given by $|y| \le rs$ and $x = rc$, for which an equation is

$$f(x, y) = \left||rs - |y|| - (rs - |y|)\right| + |x-rc| = 0$$

So I think the octagon is given by

$$f(|x|,|y|) \ f(|y|,|x|) \ f\left(\frac{|x|+|y|}{\sqrt{2}}, \frac{|x|-|y|}{\sqrt{2}}\right) = 0$$  

To get a general formula for a regular polygon of radius $r$ with $n$ sides, denote $c_n = \cos(\pi/n)$, $s_n = \sin(\pi/n)$ and

$$f_n(x+iy) = \left||rs_n - |y|| - (rs_n - |y|)\right| + |x-rc_n|$$

then your polygon is given by

$$\prod_{k = 0}^{n-1} f_n\left(e^{-\frac{2 i k \pi}{n}} (x+iy)\right) = 0$$

Depending on $n$, you can use symmetries to lower the degree a bit (as was done with $n = 8$).
