general-topology
Can the idea behind the fundamental polygon of gluing instructions for edges be extended to 3-manifolds by gluing faces of a cube together?
Specifically, take some fundamental polygon, "extrude it" and introduce some equivalence relation for the faces. Would this cube be homeomorphic to a fibration of $[0,1]$ or $S^1$ over the surface described by the original polygon? Does this process have a name?
Perhaps the simplest interesting example is the quotient of $[0,1]$ obtained from the equivalence relation $E$ whose equivalence classes are the singletons $\{x\}$ for $0<x<1$ and the doubleton $\{0,1\}$. This identifies the endpoints $0$ and $1$ to a single point, and the quotient space is homeomorphic to $S^1$, the circle. Taking $S^1$ to be specifically the unit circle in the plane, one homeomorphism is the map
$$h:[0,1]/E\to S^1:p\mapsto\begin{cases} \langle\cos2\pi x,\sin2\pi x\rangle,&\text{if }p=\{x\}\\\\ \langle 1,0\rangle,&\text{if }p=\{0,1\}\;. \end{cases}$$
The quotient topology is exactly the one that makes the resulting space ‘look like’ the original one with the identified points glued together.
(This is really just the beginnings of an answer, because I’m not sure exactly what you want to know.)
The Gauss-Bonnet theorem is certainly your friend here. If the angle sum around a vertex is $\alpha$ then there is a point mass of Gaussian curvature of magnitude $2 \pi - \alpha$ at the vertex. Thus, if every vertex of a polyhedron with $n$ vertices has the same $\alpha$, then Gauss-Bonnet tells us that $n (2 \pi - \alpha) = 4 \pi$, and thus $$ n = \frac{4 \pi}{2 \pi - \alpha}.$$
Thus we know the number of vertices. However, this does not tell us the number of faces unless we have more information about the polyhedron - for example, the cuboctahedron and truncated tetrahedron both have 12 vertices each with angle sum $5\pi/3$; but they have different face and edge counts.
Best Answer
It sounds like you are simply asking for the concept of a fundamental polyhedron: start with a polyhedron, and then give instructions for gluing faces in pairs, using gluing maps which respect vertices and edges.
For example, the 3-torus $S^1 \times S^1 \times S^1$ can be obtained from the fundamental polyhedron $[0,1] \times [0,1] \times [0,1]$ by gluing opposite faces by translation: $(x,y,0) \sim (x,y,1)$; and $(x,0,z) \sim (x,1,z)$; and $(0,y,z) \sim (1,y,z)$. This example fits your idea of "extruding".
For a famous example which does not fit your "extruding" idea, start with a dodecahedron, which has 12 pentagonal sides. Arrange those sides in opposite pairs. Viewing each opposite pair with a birds-eye view, map the nearest face to the opposite by pushing it downwards and giving it a $\frac{1}{10}$ clockwise rotation. The result is a 3-manifold called the Poincare homology sphere, said to be the counterexample which led Poincare to correct his wrong conjecture (every closed 3-manifold with the homology of the 3-sphere is homeomorphic to the 3-sphere) to his right conjecture (every closed 3-manifold with the homotopy type of the 3-sphere is homeomorphic to the 3-sphere).
Unlike gluing side pairs of a fundamental polygon, which always produces a 2-manifold, gluing side pairs of a fundamental polyhedron is not guaranteed to produce a 3-manifold: the image under gluing of a vertex might not have the correct local topology for a 3-manifold.