The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets a large jump in the value of $F(n)$ 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][1]:

> **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.

  [1]: http://groupprops.subwiki.org/wiki/Higman-Sims_asymptotic_formula_on_number_of_groups_of_prime_power_order