Python, ^{262 \$\cdots\$ 250} 248 bytes

import math def f(l): n=len(l) if n<2:return 1 R=range p=sorted([sum((l[i][k]-l[j][k])**2for k in R(len(l[0])))for i in R(n)for j in R(n)]) p=[x//p[n]for x in p] d=len(f"{n:b}")-1 e=[] for i in R(d+2):e+=[i]*2**d*math.comb(d,i) return p==e 

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Very verbose, and may have some bugs, but here's something to kick us off.

How

Calculates the square of the distances between all possible pairs of points (including self-pairs) and normalise them by the first non-zero length. For a cube we should then see a pattern of integers \$i = 0\dots d\$ occurring \$2^{d}{d \choose i}\$ times where \$d\$ is the dimension of the cube.