Expected values of Kolmogorov complexity in a random sample
Kolmogorov complexity is only determined up to some additive constant, so it is not possible to give an exact answer. The bound that I describe here is even weaker.

Of course the expected number can be computed easily once we know how many of the $2^n$ strings have complexity less than $n_0$, so let me answer this. It is usually the first statement about Kolmogorov complexity that this number is at most $2^{n_0}-1$ - since there are only these many strings of smaller length. On the other hand, if your program says "of length $n$, take the $x$th number", then you get $2^{n_0-K(n)-C}$ strings of complexity less than $n_0$, where $K(n)$ is the prefix-free version of Kolmogorov complexity of $n$ (so at most $\log n+\log^* n + O(1)$). In more detail, the string first contains the description of the Turing machine that taken input $px$, where p is the description of a prefix-free program that outputs $n$, outputs the $x$th number of length $n$, which is $O(1)$ bits, and then this is followed by $px$.

Probably it is possible to improve these bounds, but I doubt that you could get an exact answer.