Classify the set of all TMs whose languages from the accepting problem
As suggested in the comments, the extended version of Rice's theorem clearly states this language is not in $RE$. Nevertheless, let us prove this claim via a direct reduction from a language known not to be in $RE$, let's say 
$$ \overline{HP} = \{\langle M,w\rangle \mid M \text{ doesn't halts on }w \}$$

We will show that $\overline{HP} \le L$ and conclude that $L \notin RE$.

The reduction will assume we have a machine that recognizes $A_{TM}$ (call it $R$) and goes as follow. Given an input $\langle M,w\rangle$ we construct the output string $\langle M_w\rangle$ which is a machine that, on input x, does:


repeat the loop:
1.1 run one step of $M$ on $w$
1.2 run one step of $R$ on $x$
1.3  if 1.1 halts - the machine $M_w$ rejects. If 1.2 accepts - the machine $M_w$ accepts.


It is easy to verify this is a computable reduction. Let's just verify it is valid.

Case I: $\langle M,w\rangle\notin \overline {HP}$, then eventually (say after $T$ steps)  $M$ will halt on $w$ before the computation of 1.2 concludes, thus $L(M_w)$ can be decided in less then $T$ steps and in particular it is decidable. Therefore $A_{TM} \not\le L(M_w)$ and thus $\langle M_w\rangle \notin L$.

Case II: $\langle M,w\rangle\in \overline {HP}$, then $M$ never halts on $w$, which means that only step 1.2 is relevant, which means that $M_w$ behaves in this case just like $R$. So it holds that $L(M_w) = A_{TM}$ and in particular, $A_{TM} \le L(M_w)$. Then, $\langle M_w \rangle \in L$.