This is just a partial answer giving a solution to a dual problem. I will use the language of toric varieties (http://www3.amherst.edu/~dacox/), but in essence this answer is purely combinatorial.
I will construct a smooth 3-dimensional toric variety whose fan has combinatorial structure of icosahedron. Unfortunately, I am not sure that this toric variety is projective (Update David proves in the comment below that this variety is not pojective!*). If it were projective this would of course give a solution to your question but I have doubts that it is not projective (this should not be hard to check). But even if we can not get the dodecahedron this way, we will get a collection of Delzant polytops quite "close" to dodecahedron it terms of thire combinatorial structure.
In terms of combinatorics, fist thing what we will do is the following: we will show how to decompose $\mathbb R^3$ into $20$ rational simplicial cones, where each simplicial cone can be sent to the positive octant by a matrix from $SL(3,\mathbb Z)$. There will be in total $12$ rays and each ray is in the border of $5$ simplicial cones (just like for icosahedron).
Construction. First we need to chose an integral lattice $N$ in $\mathbb R^3$, it will be an index two sublattice in $\mathbb Z^3$; $(a,b,c)\in N$ if $a,b,c\in \mathbb Z$, $a+b+c\in 2\mathbb Z$. Next we specify $12$ points in $N$ that lay on $12$ ray of our fan. These are: $(\pm 1, \pm 1,0)$, $(\pm 1, 0, \pm 1), (0,\pm 1, \pm 1)$. In fact these points are vertices of the cuboctahedron http://en.wikipedia.org/wiki/Cuboctahedron . Cuboctahedron has $8$ triangular faces and $6$ square faces, and to finish the construction we should cut each square face by a diagonal into two triangles. This is done as follows: the faces $z=\pm 1$ are cut into two by the plane $x=0$, faces $y=\pm 1$ by $z=0$, and the faces $x=\pm 1$ by $y=0$. Now it is not hard to see that the obtained triangulation of cuboctahedron gives us a decomposition of $\mathbb R^3$ into $12$ standard simplicial cones (just notice that the triples of vectors ($(1,1,0)$, $(0,1,1)$, $(1,0,1)$) and $((0,\pm 1,1), (1,0,1))$ from integral bases in $N$).
Now we can ask the question. Have we constructed the fan of a projective variety? One can answer this question (but I don't do it here (Update David proved in the comment that this example is not projective)). First of all, if we don't cut square faces of cuboctahedron by diagonals, we have a fan of a singular projective variety with $6$ ordinary double singularities (i.e. singularities given locally by $(x^2+y^2+z^2+t^2=0)$). A moment polytope of this variety is dual to cuboctahedron and is called rombic dodecahedron http://en.wikipedia.org/wiki/Rhombic_dodecahedron . This is not a Delzant polytope because it has $8$ bad vertices. Now, this polytope has $12$ faces and in order to make the polytope Delzant we should just generically perturb the faces (by replacing them by nearby parallel planes). Any such generic perturbation will give us Delzant polytope. One just need to check if among these polytopes there will be the Dodecahedron... In other words, each perturbation corresponds to a symplectic structure on a small resolution of the singular variety, but it is not clear we can get the resolution corresponding to choices of diagonals in squares that we made.
