It is proved in

Holt, D. F., Enumerating perfect groups. J. London Math. Soc. (2) 39 (1989), no. 1, 67–78

that

$n^{2l(n)^2/27−dl(n)} \le F(n) \le n^{l(n)^2/6+l(n)}$

for some constant $d$, where $l(n) = \log_2(n)$. The lower bound is coming from Higman's construction of large numbers of $p$-groups of class 2, and the general belief seems to be that the lower bound is close to being the correct number.

I think that there might be better results known now. You could try searching publications of Laszlo Pyber, but I don't have time right now!

It is proved in

Holt, D. F., Enumerating perfect groups. J. London Math. Soc. (2) 39 (1989), no. 1, 67–78

that

$n^{2l(n)^2/27−dl(n)} \le F(n) \le n^{l(n)^2/6+l(n)}$

for some constant $d$, where $l(n) = \log_2(n)$. The lower bound is coming from Higman's construction of large numbers of $p$-groups of class 2, and the general belief seems to be that the lower bound is close to being the correct number.

I think that there might be better results known now. You could try searching publications of Laszlo Pyber, but I don't have time right now!

Added later: Nick's answer is much more accurate than mine. But, in case it is of any interest, let me add that the main result of the paper I mentioned was an estimate of the number ${\rm perf}(n)$ of finite perfect groups of order at most $n$, which (perhaps surprisingly) is also large, and satisfies:

$n^{l(n)^2/108−cl(n)} \le {\rm perf}(n) \le n^{l(n)^2/48+l(n)}$, where $c=11/36$.