The right answer is 52! divided by all of the symmetries. (with some minor corrections for the hands that are fixed under a given symmetry).
What are all of the symmetries?
Well you can choose to swap red and black, and then independently swap the two red and black suits. That's a factor of 8. In the standard configuration of freecell there are 8 piles, of which 4 have 6 cards and 4 have 7. So that is another factor of 24*24.
So the answer for the number of possible freecell games is approximately 52!/8/4!/4! ~ 1.75e64. Which in order of magnitude is a lot closer to 52! (8.07e67) than 13! (6.323e9).
Note that I say approximately because I did not calculate the correction terms for the freecell games which remain the same under some of those symmetries. This will increase the number of possible freecell games somewhat, but the correction will be relatively minor.
Cheers,
Ben