If each face of a die is a normal n-sided polygon, how many faces does the die have? A die with normal 4-sided polygon faces has 6 faces. 
A die with normal 5-sided polygon faces has 12 faces.
If a die has normal $n$-sided polygon faces, how many faces does it have?
Is there a general formula for this, and is there a rigorous proof?
Intuitively, if $n$ is odd, then the number of faces is $2n+2$, but I can't think of a rigorous proof for this. The hard cases seem to be when $n$ is even.
 A: Most relevant aspects have already been discussed in comments.
The most typical generalization of a “die” to other faces would be a platonic solid. There are exactly five platonic solids, including the dodecahedron with its pentagonal faces which you already mentioned in the question. Three of these have the same face shape, namely the tetrahedron, the octahedron and the icosahedron. They all have triangular faces, but four, eight or twenty faces respectively.
So at this point you can already see that there can be no formula to deduce the number of faces from the number of corners per face alone.
In my opinion the most useful formula you can get for computing the number of faces is Euler's polyhedron formula which states that for each convex polyhedron you get $$V-E+F=2$$ where $V$ is the number of vertices, $E$ the number of edges and $F$ the number of faces.
In a comment you asked why there are only five platonic solids, but that is probably better discussed in the question Platonic Solids. Two of the current answers there make use of Euler's polyhedron formula, too.
One could generalize the question further, e.g. by dropping the convexity constraint (and therefore considering Kepler–Poinsot polyhedra as well) or by dropping one of the transitivity requirements of regular polyhedra. For example, the ten-sided die is convex and face-transitive but not edge- or vertex-transitive, and its faces are not regular polygons. So it all depends on what you call a “die”.
