A good question!
I think the current situation is, nobody knows.
I doubt that the number can be calculated, some sort of enumeration is required.
A brute-force approach is out of the question - one would need to test all the ED grids (5,472,730,538), but also the 3,359,232 variants for every case, since these transformations don't preserve the AK property. We can only reduce this by a factor of 4, for rotation/reflection symmetries, but still the number of grids to test is enormous.
But these AK grids are likely to be very rare, since up to 8 extra eliminations can be associated with each cell (the central square has the full 8). Rarity seems to be confirmed by a random grid generator that tested over 1.5 billion grids before finding one anti-knight case … not very scientific, but it does suggest that an exact count is probably easily obtained ...