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.
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 $u=10+4\sqrt{5}$.