I found this:
The tesseract is a four dimensional cube. It has 16 edge points $v=(a,b,c,d)$, with $a,b,c,d$ either equal to $+1$ or $-1$. Two points are connected, if their distance is $2$. Given a projection $P(x,y,z,w)=(x,y,z)$ from four dimensional space to three dimensional space, we can visualize the cube as an object in familiar space. The effect of a linear transformation like a rotation $$ R(t)=\pmatrix{1&0&0&0\\0&1&0&0&\\0&0&\cos(t)&\sin(t)\\0&0&-\sin(t)&\cos(t)} $$ in $4D$ space can be visualized in $3D$ by viewing the points $v(t) = P R(t) v$ in $\mathbb R^3$.
So how does $P$ actually look?
The particular $P$ used to make your animation loop looks like projection away from a point directly above the the $4$-cube. Intuitively, put a light source at $\mathbf{c} = (0, 0, 0, h)$ and map each point $\mathbf{x} = (x, y, z, w)$ with $w \neq h$ to its "shadow", the intersection of $\mathbf{R}^3 \times \{0\}$ with the ray from $\mathbf{c}$ through $\mathbf{x}$: $$ P(x, y, z, w) = \frac{h}{h - w}(x, y, z). $$
If we assume the $4$-cube has edges of length $2$ and is centered at the origin (i.e., it's the $1$-skeleton of the $4$-fold Cartesian product $[-1, 1]^{4}$), then $h \approx 3$: When the $4$-cube is aligned with the coordinate axes (so the projection looks like a small cube nested in a large cube, with corresponding vertices joined), the "top face" looks about twice as large as the "bottom face".