There are 
$$ \frac{\prod_{i=0}^{k-1}(n-i)^2}{k} $$
possible directed cycles with $2k$ vertices.  Each such cycle occurs with probability $n^{-2k}$, so the exact expectation of the number of cycles is
$$ \sum_{k=1}^n \frac{\prod_{i=0}^{k-1}(n-i)^2}{kn^{2k}}
 = \sum_{k=1}^n \frac{\prod_{i=0}^{k-1}(1-i/n)^2}{k}. $$
As $n\to\infty$, that sum is asymptotic to the integral
$$ \int_1^\infty \exp(-x^2/n)/x~dx
 = \left[ -\frac12 Ei(1,x^2/n) \right]_1^n = \frac12\log n + O(1),
$$
where $Ei$ is the exponential integral function and I'm relying on Maple a bit.