The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets much larger values for $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$.
The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:
Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$
(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:
Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{2/27+o(1)\mu^2}$$ as $\mu\to\infty$.
The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:
Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.
An interesting extra tidbit from that paper is the following:
Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).
(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result!
$(\dagger)$ I said this was a conjecture earlier - my mistake.