general-topology
$n \mathbb{T} = \mathbb{T} \# \mathbb{T}\# \cdots \# \mathbb{T} =$ a sphere with $n$ handles (connected sum of $n$ tori $\mathbb{T}$)
If we encircle a wire frame cube with a small tube, we get a connected sum of how many tori $\mathbb{T}$?
What about if we do the same thing to a tetrahedron and a dodecahedron?
@DanielRust is correct; what you have constructed is just another compact manifold whose boundary is two genus-2 surfaces, not the cylinder on that surface. Here's some more detail.
$Z$ is homotopy equivalent to the two cylinders joined by line segments rather than disks. More precisely, $(X \coprod X' \coprod I_0 \coprod I_1) / R$, where each $I_j$ is a unit interval and the relation $R$ glues $0 \in I_j$ to $(t, j) \in X$ and $1 \in I_j$ to $(t, j) \in X'$ (choosing some base point $t \in T$). Using another homotopy equivalence to collapse $X$ and $X'$ to $T$ and $I_0$ to a point, we get $T \vee T \vee S^1$.
To show that $\Bbb{Z}^2 \ast \Bbb{Z}^2 \ast \Bbb{Z}$ is different from the fundamental group of the genus-2 surface, we'll need something like MO "Free splittings of one-relator groups".
Note that your construction works similarly if you replace $T$ by any manifold $M$, and some simpler examples may be helpful. If you take $M$ to be $S^1$, the result is a twice-punctured torus; if you take $M$ to be $S^2$, the result appears to be a twice-punctured $S^2 \times S^1$.
That's because not all the points are identified. In fact, they are divided into two groups of five, each of which is identified as a separate point. See the diagram below and try to chase them.
Best Answer
Intuitively: deform the "top" and "bottom" of the cube into circles. Remove two opposite vertical edges. Then the wireframe will clearly be a connected sum of three tori. But adding in those two missing edges is like adding one of the missing edges is clearly like adding a "diameter" to one of the torus components, and this is just adding another torus.
Slightly more formally, start with the "tubed" base square, and add edges one by one. If you count how many times you needed to add a handle, as opposed to just stretching out an "appendage", you will come up with four, which makes five in total. In fact, you could start with just a single vertex. Then the number of times you'd have added a handle would be exactly five.
It's easy to see that you added a handle at those times when you added an edge without also adding a vertex. That should tell you what the general formula for the number of handles is. You can verify that it works in case of tetrahedron (by hand, as in case of the cube). For dodecahedron, you'd probably better stick to applying it.