TL;DR: $n \centerdot 2^n \centerdot n^{mn}$
where $\mid Q\mid$ $=$ $n$ and $\mid \Sigma\mid$ $=$ $m$.
We'll go through each element of a DFA 5-tuple to figure out the various combinations that would each yield a unique DFA. The 5-tuple consists of ($Q$, $\Sigma, \delta$, $s$, F)
$s$:
Any 1 element of $Q$ can be the start state. Thus there are $\mid Q\mid$ = $n$ ways to choose $s$.
F:
Any number of elements of Q can be accept states, therefore all subsets of Q are valid choices for F. The number of possible subsets for a set of cardinality n is 2$^n$. Another way to say this is the cardinality of $Q$'$s$ power set $P(Q)$ is 2$^n$
$\delta$:
$\delta$ is defined as f: $Q$ x $\Sigma\rightarrow Q$ i.e. the $\delta$ function's domain is $Q$ x $\Sigma$ and its range is $Q$. The cardinality of the domain is $mn$ where $\mid Q\mid$ $=$ $n$ and $\mid \Sigma\mid$ $=$ $m$, and that of the range is $\mid Q\mid$ $=$ $n$. Thus there are $n^{mn}$ ways to choose $\delta$.
$Q$ and $\Sigma$ are given; there is only $1$ way to choose them.
So the total number of ways to choose amongst the 5 elements of a DFA where $\mid Q\mid$ $=$ $n$ and $\mid \Sigma\mid$ $=$ $m$ is
$n \centerdot 2^n \centerdot n^{mn}$
Better 5 years late than never, huh?