Connected sum of $n$ tori $\mathbb{T}$ $n \mathbb{T} = \mathbb{T} \# \mathbb{T}\# \cdots \# \mathbb{T} =$ 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?
 A: 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.
A: Very generally, suppose you have a finite connected graph $G$ (which may have loops and multiple edges).  Replace each vertex of $G$ by a sphere with $n$ disks removed, where $n$ is the number of edges coming out of the vertex, and attach cylinders between these punctured spheres according to the edges.  This gives a closed connected $2$-manifold which I'll call $T(G)$.
Now recall that the Euler characteristic of $G$ is $\chi(G)=V-E$, where $V$ is the number of vertices of $G$ and $E$ is the number of edges.  If you have an edge of $G$ whose endpoints are distinct, consider the graph $G'$ obtained from $G$ by deleting that edge and identifying its endpoints together.  Note that $G'$ is still connected and that $\chi(G)=\chi(G')$, because we have removed one edge and one vertex.  Furthermore it is easy to see that $T(G')$ is homeomorphic to $T(G)$ (this is just the fact that if you take two spheres, cut a hole out of each and connect them by a cylinder, then you get a sphere again).
We can delete edges like this over and over until there are no more edges left to delete.  If there are no more edges we can delete, that means that all the edges must be loops; since all the graphs we get in this process are connected, this means we've reached a graph $H$ with only one vertex.  But if $H$ has just one vertex and $g$ edges (which are all loops at that vertex), it is clear that $T(H)$ is just a connected sum of $g$ tori.  We can also calculate that $\chi(H)=1-g$.
To sum up: given any finite connected graph $G$, there graph $H$ with one vertex which has the same Euler characteristic as $G$, and $T(G)\cong T(H)$ is a connected sum of $g$ tori, where $g$ is the number of edges in $H$.  Since $\chi(G)=\chi(H)=1-g$, we can restate this as follows: $T(G)$ is a connected sum of $1-\chi(G)$ tori.
In the specific cases you ask about, $G$ is the $1$-skeleton of a cube, a tetrahedron, or a dodecahedron.  Let's answer the question more generally when $G$ is the $1$-skeleton of a convex polyhedron with $F$ faces.  In that case, by Euler's formula for convex polyhedra, we know that $F+\chi(G)=2$.  We conclude that $\chi(G)=2-F$, so $T(G)$ is a connected sum of $F-1$ tori.
