EDIT
Unfortunately, the details will not be ready until after the bounty expires. Someone else may be able to use the sketch below.
Let S be the multiset of pieces used at the start of a chess game. Let Q=(64_S) be the number of possible arrangements, one piece per square, using S on a chessboard. Q is (the value of) a multinomial coefficient using data from S, and is about 4x10^42.
If one looks at the number of possible arrangements using a legal submultiset S' of S (in particular the collection of pieces arising from a legal game of chess involving no promotions), one encounters an expression like Q(1 + 30/33*(1 + 29/34*(1 + 28/35...))), which evaluates to something less than 5Q. Further, considering legal collections T, one needs only those submultisets of T which are legal and have not been considered earlier in a sequential enumeration of legal submultisets. Thus, a good approximation to the total number of all positions considered should be the sum over T a maximal legal multiset of terms (64_T).
The T I have found so far that gives a maximum value for (64_T) is slighly less than 8Q, involves one pawn capture and 3 queen promotions of different colors. After two pieces being captured, it is hard for me to see any optimal T occurring.
There are 165 possible multisets arising from promoting 8 pawns of one color. I suspect an upper bound for the number of maximal legal multisets is 165165, given that 2k promotions usually require k captures. This suggests an upper bound of 272258Q for the number of all possible positions, or about 217800Q.
I was hoping to find a T such that I could beat Joel Hamkins's ratio of legal game positions using T/(64_T) while also satisfying (64_T) > Q, but time has run out for that. It looks like 2^-32 remains unverified at this writing.
TIDE
If the idea is to come up with a rough figure, the following approach using labeled pieces might help.
Consider the white king's pawn. It cannot inhabit 24 of the squares on the chessboard (I assume promotion is strict and monotone. I let others argue whether to add 8 to the number 24.) . For the other king side pawns 24 grows to 26, 30, and 36. Starting with these ratios, a set of four pawns labeled with their starting squares can occupy about 9% of all possible positions available. Raise this to the fourth power, and one gets a ratio of 6x10^-5 as an upper bound on positions of labeled pawns, where as a rough count I include pawns sharing the same square.
Now the bishops provide another factor of about 6x10^-2, the two kings together a factor of about 7/8, and the rest provide a factor that is likely closer to 1. This gives a weak upper bound for all 32 labeled pieces of 3 x 10^-7 fraction of legal positions to all positions.
I imagine with refinements one can probably subtract one more from the exponent. Hopefully the fraction involving unlabeled pieces will not be much different.
I just noticed domotorp's comment regarding positions having just 28 pieces. While this suggests the fraction of legal positions may be higher, for the labeled case, the ratio (all possible for 28) to (all for 32) should still be small, and should have a small effect on the rough calculation.
And now I just noticed more recent comments from Joel David Hamkins, who made some similar and more refined observations. Perhaps he can place the above musings on a more rigorous footing, and maybe someone can convert the labeled analysis to an unlabeled one.