If we "wrap" coordinates on plane we get torus surface, what we get when wrap coordinates in cube? In computer games, we can often see worlds that are like finite planes, whose opposite edges are stitched together (if you go up, you end in bottom, if to left - on the right). The surface we got appears to be a torus.
I am wondering, what we get, if we go to 3d, "wrap around" coordinates in cube? I think, we should get some figure in 4D, however, I broke my mind when trying to imagine in my head what we get.
Do anybody know what is the figure, how to find it and imagine it?
EDIT:
I have found this image on the Internet:

Is figure on the right ("torus" with 3 holes) that figure I am searching? I guess it's not, because on the left inner figure does not contain edges of cube.
 A: Begin with $1$ dimension: what do you get when you join the endpoints of the line segment $[0,1]$? You get a circle, $S^1$.
When you join the opposite edges of $[0,1] \times [0,1]$ you get, as you say, the "flat" torus, $S^1 \times S^1$.
It is easy then to understand what happens in general: the $n$-dimensional cube $[0,1] ^n$ has $n$ pairs of opposite "faces" (which, in turn, are $n-1$-dimensional cubes). When you identify these opposing faces you will get an $n$-dimensional torus, $(S^1)^n = \underbrace{S^1 \times \dots \times S^1} _{n \text{ times}}$.
A: 
what we get, if we go to 3d, "wrap around" coordinates in cube?

It's called a 3-torus.
The 2-torus is already a bit tricky, because the 2-torus from computer screens is flat, whereas the usual depictions of 2-tori in 3-space are not flat.  You can embed a flat 2-torus isometrically$^1$ (i.e. angles and distances on the torus remain the same) in Euclidean 4-space, but that's not easy to imagine.
With the 3-torus it's even harder, but shares some similarities with the 2-torus:  it's finite and has no borders.  And it can be assigned a geometry that's flat everywhere.
A good way to get a better understanding is imagining how one moves around in such a space.  And you can have different types of loops:  A "small" loop can be smoothly contracted to a point.  But when you have a loop that starts in the middle, then exits at top and thus re-enters at the bottom, and finally closes in the middle again; then no matter how you move around or stretch that loop, you can never shrink it to a point without tearing.

Is figure on the right ("torus" with 3 holes) that figure I am
searching?

No.  What the image appears to explain is this:
There is a surface sitting inside $T^3$ (a 3-torus).  If that surface was sitting in an ordinary cube, it had 6 holes / openings, one on each face of the cube.  That surface is however sitting in $T^3$ so that the bottom opening connects with the top opening, dito for left-right and front-back holes.  The result is a betzel surface (3-holed object) without boundary, where the 3 gluings that happened are colored red, green blue.
A similar reasoning applies when the author considered the filled-out structure.  In that case the result is a filled bretzel which has a boundary (the bretzel surface from the previous paragraph).
The bretzel surface can be assigned a geometry that has constant negative curvature everywhere, i.e. it's "natural" geometry is hyperbolic.  And the space is finite and has no borders.  But just as with the 2-torus, embeddings in 3-space will distort it in such a way that the curvature is no more const and changes from place to place.$\def\R{\Bbb R}$

Note on embeddings of $T^n$ in $\R^m$
You cannot embed the 3-torus in 3-space: Closed $n$-manifolds do not embed in $\R^n$.  Hence a lower bound for $T^3$ is $m\geqslant4$.
Upper bound for smooth / continuous / flat (isomentric) embeddings are given by Whitney's embedding theorem and by Nash's embedding theorem. This video explains an embedding$^1$ of flat $T^2$ in $\R^3$, which is fractally crumbled.  Similar follows for $T^3$ in $\R^4$:  Nash's theorem for isometric $C^1$ embeddings$^1$ states that it's possible for $m=4$.  For a smooth, flat embedding of $T^3$ you'd need at most $30=3\!\cdot\!(3\!\cdot\!3+11)/2$ dimensions, thus $m\leqslant 30$.
There is an explicit construction of a flat, smooth embedding of $T^2$ in $\R^4$.  If similar works for $T^3$, then there is a flat, smooth embedding of $T^3$ in $\R^6$.
If the embedding should be smooth but need not be flat, Whitney's theorem gives an upper bound of $\R^5$, see here.

$^1$Nash's theorem for $C^1$ embeddings only guarantees approximate isometric embeddings, i.e. there might be an approximation error and the embedding not exactly isometric.  That error may be chosen to be arbitrarily small, though.
