LOWER BOUND: We can similarly to the notion of "k-sets" chopped off by hyperplanes, instead consider chopping using a sphere. Then the total number of sphere-cut k-sets (all k combined) is upper bounded by 2*(N choose D+1).
Actually, if you map points in D-space onto the D-dimensional surface of a sphere in D+1 dimensional space via stereographic projection, then the hyperplane-cut k-sets on the higher diml set correspond to the sphere-cut k-sets on the lower diml set.
I think the Clarkson-Shor proof still works even if points restricted to lie on a sphere. If so, then we get a lower bound for N large D fixed of form #distanceOrderings >= C_D * N^(D+1) / f(N) where f(N) grows to infinity arbitrarily slowly.
--arggh, sorry, I forgot to multiply by 2/N. So it really would yield #distanceOrderings >= C_D * N^D / f(N) where f(N) grows to infinity arbitrarily slowly. We anyhow remain stuck in a quagmire on Wilson's ordering problems in which our upper bounds are roughly the square of our lower bounds.