I am not a physicist, but I suppose it is because the s-orbital has an amplitude peak at r=0?

In other words, it is not so much the antiproton falling to the nuclear surface as much as the antiproton finding itself at the nuclear surface.

EDIT: The context is that the antiproton was in an orbital with large principal and azimuthal quantum numbers. Still, there would be some non-zero probability of the antiproton finding itself close to the nucleus, no?