In dimension $d$, we will define a Hadamard vector as a vector that (up to normalisation) has either a 1 or a -1 as every entry.
In dimensions $d=2^n$, for some natural $n$, one can find bases consisting only of Hadamard vectors. We will call such bases Hadamard bases.
How many Hadamard bases are there as a function of $n$?
For $n=2$, we can express the number of bases as (ignoring phase for the moment)
$2^4 {{4}\choose{2}}{{2}\choose{1}}^2/4! $
We can interpret this as having $2^4$ choices for the first vector in our basis; then the second must differ in sign in two of the four places; then for both the pairs of coefficients in which the first two do not differ, the third must differ in one place, and then we have only two choices (due to phase) for our final basis vector. We divide by $4!$ as we do not care about order. Then, we can divide by 2^4 if we care about uniqueness up to phase.
However this approach does not seem to be tenable for $n=3$.