In what kind of space does this object live? Let me quickly build up some background. 
One way to build a hypercube is to take cubes, and start gluing them together, face to face, such that each edge is shared by $3$ cubes. You complete the hypercube with $8$ cubes. This involves rotating cubes in $4$ dimensions, but if you forget about the geometry, you can still do this abstractly. 
Separately, lets take a torus that is made from 5 squares, like this: 

where you identify opposite sides in the usual way to make a torus.
Now, lets combine these concepts. Let's take these tori and (abstractly?) glue them together, square face to square face, such that each edge is shared by 3 tori. It turns out this only takes $6$ of these tori.
My question is this: where the hypercube lives nicely in $4$-space, is there a nice, familiar, space where this clump of tori lives? Is there a nice way to think about this kind of object?
EDIT:
As requested in the comments, here is a gluing diagram for what I am proposing:

The numbers and colors mean the same thing, they identify square faces to be glued together. The letters correspond to the vertices, so one knows the orientation of the squares to be glued.
 A: tl;dr: It lives in 5 dimensions.
First, note that the square faces are not planar. The figures you have drawn there are projections of Clifford tori. Clifford torus lives in 4 dimensions. If you project the central square back onto the Clifford torus, the vertices will have these coordinates:
$$
A=\left[\cos-\frac35\pi;\sin-\frac35\pi;\cos-\frac15\pi;\sin-\frac15\pi\right] \\
B=\left[\cos-\frac15\pi;\sin-\frac15\pi;\cos\frac35\pi;\sin\frac35\pi\right] \\
C=\left[\cos\frac35\pi;\sin\frac35\pi;\cos\frac15\pi;\sin\frac15\pi\right] \\
D=\left[\cos\frac15\pi;\sin\frac15\pi;\cos-\frac35\pi;\sin-\frac35\pi\right] \\
$$
It is obvious that the sides of the square are equal and the diagonals are equal. Let's show that the diagonal is the same length as the side, namely $||A-B||=||A-C||$. We can evaluate the sines and cosines first:
$$
A=\left[\frac14\left(1-\sqrt5\right);-\sqrt{\frac58+\frac{\sqrt5}8};\frac14\left(1+\sqrt5\right);-\sqrt{\frac58-\frac{\sqrt5}8}\right]\\
B=\left[\frac14\left(1+\sqrt5\right);-\sqrt{\frac58-\frac{\sqrt5}8};\frac14\left(1+\sqrt5\right);\sqrt{\frac58+\frac{\sqrt5}8}\right]\\
C=\left[\frac14\left(1-\sqrt5\right);\sqrt{\frac58+\frac{\sqrt5}8};\frac14\left(1+\sqrt5\right);\sqrt{\frac58-\frac{\sqrt5}8}\right]
$$
First, we calculate $||A-C||^2$:
$$
||A-C||^2=\left(2\sqrt{\frac58+\frac{\sqrt5}8}\right)^2+\left(2\sqrt{\frac58-\frac{\sqrt5}8}\right)^2=\frac52+\frac{\sqrt5}2+\frac52-\frac{\sqrt5}2=5
$$
And now $||A-B||^2$
$$
||A-B||^2=\left(\frac14(1-\sqrt5-1-\sqrt5)\right)^2+\left(\sqrt{\frac58+\frac{\sqrt5}8}-\sqrt{\frac58+\frac{\sqrt5}8}\right)^2+\left(\frac14(1+\sqrt5-1+\sqrt5)\right)^2+\left(\sqrt{\frac58+\frac{\sqrt5}8}+\sqrt{\frac58+\frac{\sqrt5}8}\right)^2=\\
=\frac54+\frac58+\frac{\sqrt5}8+\frac58-\frac{\sqrt5}8-2\sqrt{\frac58+\frac{\sqrt5}8}\sqrt{\frac58-\frac{\sqrt5}8}+\frac54+\frac58+\frac{\sqrt5}8+\frac58-\frac{\sqrt5}8+2\sqrt{\frac58+\frac{\sqrt5}8}\sqrt{\frac58-\frac{\sqrt5}8}=5
$$
So $||A-B||=||A-C||=\sqrt5$.
This means that the sides and diagonals of these squares are the same so they are not planar. Their vertices actually form regular tetrahedra. 5 tetrahedra form a pentachoron (4-simplex) and 6 pentachora form a hexateron (5-simplex) which lives in 5 dimensions.
A: Any geometric manifold can live in any kind of space, given a suitable injection. But there are two common ideas. Consider a polyhedral surface, say a cube. On the one hand it lives on the sphere $S^2$ in the sense that it is a topological decomposition of the sphere into polygons, each of which is topologically an $S^1$. But it also lives naturally in locally-Euclidean 3-spaces such as $E^3$, in the sense that it can be smoothly embedded.
Every polytope is a topological decomposition of some manifold, but not every polytope can be locally embedded in a higher space. For example every polyhedral decomposition of the projective plane $P^2$ must self-intersect as an immersion when injected into $E^3$, it cannot be smoothly embedded. On the other hand a toroidal polyhedron can be embedded in $E^3$ and will have holes through it like a donut or a pretzel.
Similarly, the hypercube is a polyhedral decomposition of $S^3$ and can be embedded in $E^4$. Your 6-torus polychoron is a decomposition of a far more complicated toroidal 3-manifold. Whether or not it can be embedded in $E^4$ would take a lot of heavy maths to discover, although I expect that somebody already has.
There is no easy road to visualising objects in higher dimensions, but treating them as manifolds and studying their topology is perhaps the most accessible.
A: I have a more straightforward answer which doesn't require any calculations.
These tori you have there are actually Clifford tori which live in 4D. Look at the vertices, each of them has degree 4 and they are connected each to each. You have to take into account the wrapped edges also. And all of these edges are the same length which means that the vertices and edges form a regular pentachoron (4-simplex). And 6 regular pentachora form a regular hexateron (5-simplex) which lives in 5D.
