Is {: L(M) ∈ NP} ∈ NP?
Your language is not even decidable.
Given a Turing machine $T$, define $M_T$ as a Turing machine that simulates $T$ on an empty input, and then decides any $\mathsf{EXPSPACE}$-complete language $L_C$ of your choice (notice that the function $f(T) = M_T$ is computable).
Notice that the language $L(M_T)$ is $L_C \not\in \mathsf{NP}$ if $T$ halts and $\emptyset$ otherwise.
Suppose now that there exists a Turing machine $M^*$ that decides $L$.
Then, you could design a Turing machine that solves the halting problem. When given input $T$, this Turing machine behaves as follows: it computes $M_T$; it simulates $M^*$ with input $M_T$; it rejects if $M^*$ accepts, otherwise it accepts.
Notice that if $M^*$ accepts then you must have $L(M_T) = \emptyset$, showing that $T$ does not halt.
On the contrary, if $M^*$ rejects, then $L(M_T) \not\in \mathsf{NP}$ implying that $L(MT)=L_C$ and hence $T$ halts.