You can get $4k$ pentagonal faces for any $k≥3$. Take $k$ pentagons meeting at a point (in the plane, or on a sphere). Between each adjacent pair of pentagons, draw another pentagon, making a second ring of $k$ pentagons. Each of the original $k$ pentagons has a single exposed edge left; to this edge and the adjacent two edges of second-ring pentagons, add two more to form a new pentagon. This makes a third ring of $k$ pentagons. Joining the $k$ new vertices of this ring to a single point completes a fourth ring of $k$ pentagons.
Having drawn the planar graph, Steinitz's theorem says such a polyhedron exists.
The initial vertex, and the final vertex, have degree $k$; all other vertices have degree three. When $k=3$ this is a standard construction of the regular dodecahedron.
The polyhedron can be made with an axis of $k$-fold rotation through the two vertices of degree $k$; we also have reflections and a half-turn taking one such vertex to the other, for at least $4k$ symmetries.
On the other hand, clearly the number of faces must be even (since 5 times the number of faces is twice the number of edges). So, the remaining possibilities for the number of faces are equivalent to 2 mod 4.
At Canonical Polyhedra. you can get the seven hexahedra and their duals. These are your 11, 2, 1, 3, XX, 7, 4. You are missing the (3,3,4,4,4,4) case. Vertices {{-0.930617,0,-1.00},{0.930617,0,-1.00},{-0.57586,-0.997418,0.07181},{0.57586,-0.997418,0.07181},{-0.57586,0.997418,0.07181},{0.57586,0.997418,0.07181},{0,0,1.81162}}, with faces {{1,2,6,5},{1,3,4,2},{1,5,7,3},{2,4,7,6},{3,7,4},{5,6,7}}}
Another view
One way to prove you have all of them is to start with the pyramid / 5-wheel graph. The pentagon with points connected to the center. A polyhedral graph is a planar graph that is 3-connected (no set of 3 vertices that disconnects the graph).
By repeated vertex splitting and merging, all n-faced polyhedra can be derived from the n-faced pyramid. You are missing the shape that merges two neighboring corners of a cube. This is Tutte's Wheel Theorem. Here is how the hexahedral graphs connect.
Canonical Polyhedra has code and pictures.
Best Answer
This is a very interesting question, and I think I understand what ipmcc has in mind. There's a lot of Mathematica code involved in generating these images, so I'll put that at the end.
The first thing to understand is "boards are halfway between being coplanar with the 2 faces on either side". If I understand things correctly, the placement of a board with respect to the dodecahedron should look something like this:
With all of the boards, we should have one new face for every old edge. "What polyhedron has a face for every edge of an original one?" has been asked and answered in Is there such a thing as the "edge-face dual" of a polyhedron, and is the "edge-face dual" of a cube a rhombic dodecahedron?: It's the dual of the rectified polyhedron. A rectified dodecahedron is an icosidodecahedron, and the dual of that is a rhombic triacontahedron (with 30 golden rhombus faces). If we draw a triacontahedron on the outside of our dodecahedron, we get something like the following:
The problem is that that doesn't really look anything like "conceptually close" to the link @TonyK posted in a comment. We don't want the maximum possible piece of each board (a golden rhombus), we just want a sliver of each rhombus around the edges of the original dodecahedron. I think the intent was for something that looks like this (possibly with less wide slivers):
However, the hexagons in that image are infinitely thin, whereas the question referred to "woodworking" and "beveling"; we need to thicken them up. We can't just thicken them naively because of the dihedral angles. They need to be frusta: beveled so they're part of a hexagonal pyramid with "top vertex" at the center of the figure. It should look something like this:
By symmetry, you only need to build 30 copies of the same frustum. Fixing a length for the long diagonal, a frustum depends on two parameters. Let t be a parameter ranging from 0 to 1 where 0 means you have a whole golden rhombus face and 1 means you have just a line segment that is an edge of the dodecahedron (interpolating linearly). Let s be a parameter ranging from 0 to 1 where 0 means you have the whole pyramid and 1 means you have just the hexagon (interpolating linearly). Then the image with the flat hexagons had t=0.85 and s=1, and the image with the solid frusta had t=s=0.9. To see the effects of changing t and s qualitatively, here is a Youtube animation of the whole dodecahedron as t and s are changed, and here is a Youtube animation of just one frustum.
In sloppy Mathematica notation, where
phiis the golden ratio $\varphi=\frac{1+\sqrt{5}}{2}$, andsandtare the parameters just described, the following is a list of faces of a single frustum (of long face diagonal length $\frac{2}{\varphi}$), where each face is a list of vertices (in a right-hand-rule order around the face), and each vertex is is a triple of x, y, z coordinates:The angles of the outer hexagon are $\tan^{-1}2\approx 63.43^{\circ}$ and $720^{\circ}-2\tan^{-1}2\approx 148.28^{\circ}$. The dihedral angle between the larger hexagon and one of the four smaller (assuming t near 1) trapezoidal faces is exactly $72^{\circ}$. However, the dihedral angle between the larger hexagon and one of the two bigger (assuming t near 1) trapezoidal faces depends on t. It's $\approx86.07^{\circ}$ for t=0.9, and in general is given by $$\cos^{-1}\left(\frac{\left(3-\sqrt{5}\right)\left(1+\sqrt{5}\right)\left(1-t\right)}{4*\sqrt{t^2+\frac{1}{2}\left(3-\sqrt{5}\right)\left(1-t\right)^{2}}}\right)$$
If you happen to have Mathematica 7 or above (the code might also work in 6), here is most of the code (with some scaffolding removed and some comments added) for generating the images and formulae referenced above. I make no promises about elegance or efficiency.