As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers.

[![polyhedron][1]][1][![projection][2]][2]

Then the odds of a triangular, square or pentagonal roll overall are
$$14.4\%,\ 50.4\%,\ 35.1\%$$
and the odds for an individual triangular, square or pentagonal face are
$$0.72\%,\ 1.68\%,\ 2.93\%.$$

This is Simpson's method.  It also represents the result of throwing the die high in the air above an adhesive surface, so that the die is well-randomized, and then after touching the surface falls quickly onto the nearest side.

The angles do not have nice exact formulas, but they are easy enough to calculate in Mathematica:

    polyhedron = PolyhedronData["SmallRhombicosidodecahedron"]
    data       = polyhedron[[1]]
    vertices   = data[[1]]
    faces      = data[[2,1]]
    triangle   = faces[[20]]
    square     = faces[[50]]
    pentagon   = faces[[62]]
    center[x_] := Table[vertices[[x[[i]]]], {i, 1, Length[x]}] // Mean // Simplify
    angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/
        (Norm[a] Norm[b] Norm[c] + (a.c) Norm[b] + (b.c) Norm[a] + (c.a) Norm[b])]
    angleof[face_] := angle[center[face], vertices[[face[[1]]]], vertices[[face[[2]]]]]
        *Length[face]
    angles = Map[angleof, {triangle, square, pentagon}] // Simplify
    N[{20, 30, 12} angles/(4 Pi)]
    N[angles/(4 Pi)]
    (angles /. Sqrt[5] -> (u - 10)/4) // Simplify // TeXForm

This gives the following angle measures for each triangular, square or pentagonal face:
$$6 \arctan \left(\frac{u-4}{24 \sqrt{5u+5}+58 \sqrt{u+1}+9 \sqrt{9
   u-3}+4 \sqrt{45 u-15}+\sqrt{3723 u-3903}}\right),\\
8 \arctan\left(\frac{1}{u+\left(2+\sqrt{5}\right) \sqrt{u+1}}\right),\\
10 \arctan\left(\frac{1}{\sqrt{10u}+3 \sqrt{u+1}}\right)$$
where $u=10+4\sqrt{5}$.


  [1]: https://i.sstatic.net/Cu8qsm.png
  [2]: https://i.sstatic.net/mInPym.png