This number is so incomprehensibly large that it probably has not been computed in the general case. For only 3 pieces left (if you're considering the kings to be a piece), then the only possible checkmates are if the 3rd piece is a rook or queen. If the 3rd piece is a rook, the mated king must be in a corner square and the rook/king have to have a specific placement, so that is an easy count. If the 3rd piece is a queen, then there are a bunch of subcases to consider, but it is certainly doable.
When you go to 4 pieces, the math already becomes extremely tedious. The first subcase is that one of the pieces is a rook or queen, and you are only mating with that piece while the other is not involved in the mate at all (the other piece can be either the winner's or the loser's, adding to the complexity). Then, you have the cases where the rook/queen is mating while defended by the other piece, and the king isn't involved. Then you have the case where the king, queen/rook, and other piece are all involved. That in of itself is a ton of cases. along with that, it is possible to mate with bishop/knight, so you then have to consider those subcases (though, like in the king+king+rook case above, the mated king needs to be cornered so there are very few cases here). I believe you can mate with pawn/knight and pawn/bishop as well, though most people in that situation would just promote to a queen (one does need to verify that this is possible because I'm not sure). One can also mate with 2 bishops. Finally, you can mate with two pawns if the king is on the back rank, which yields a few more cases.