I have the following language:
$$ L=\{ \langle M, n \rangle \mid \forall x\in \Sigma^*\ s.t\ |x|=n, M\ doesn't\ reject\ x\} $$
Where $M$ is a TM with finite $\Sigma$ and $\Gamma$.
I think that $L$ is not in $RE$, but I don't know how to prove it. I am thinking about reduction from $\overline{HP}$, but not sure if this is the right way and if it is, then how to do the reduction.
I'll assume that $M$ is intended to be a Turing machine (as opposed to, for example, just a finite state automaton). I'll also assume that "$M$ rejects $x$" means that when $M$ runs on input $x$, it enters a rejecting state (as opposed to meaning, for example, just that $M$ doesn't accept $x$). Finally, I'll assume that $\Sigma$ is a finite alphabet.
Under these assumptions your set is r.e. A semi-decision algorithm works as follows: On input $\langle M,n\rangle$, it operates parallel (interleaved) runs of $M$ on all the finitely many inputs $x$ of length $n$. If and when all these runs have rejected, then the algorithm accepts $\langle M,n\rangle$.