How to efficiently check if a superoperator is Lindbladian?
Depending on what you need this is only a (very) partial answer to the question, because it includes only necessary conditions for an operator to be Lindbladian, not sufficient ones (as far as I know).
Lindbladians have two important properties: they generate (1) trace-preserving and (2) completely positive evolution.
A necessary condition for a superoperator $\mathcal{L}$ to generate a trace-preserving evolution is $$ \mathcal{L}^\dagger[\mathbb{1}] = 0.$$ It is easy to check since it requires only Hermitian conjugation of the matrix $\mathcal{L}$ and multiplication with the identity "vector" $\mathbb{1}$.
A necessary condition for $\mathcal{L}$ to generate a positive semi-group is that the real part of the eigenvalues of $\mathcal{L}$ are non-positive. This is less easy to check since one has to diagonalise an $N\times N$ matrix.
First, note that if we allow the Hamiltonian $H$ to have an anti-Hermitian part which is a positive semi-definite operator, $H-H^\dagger \ge 0$, then $\mathcal{L}$ still generates CP time evolution $e^{t\mathcal{L}}$; it's just not trace preserving unless $H-H^\dagger=0$. Let's call the not-necessarily-trace-preserving class of superoperators Lindbladian, and call the trace-preserving subset completely dissipative. (The latter name is the terminology that Lindblad originally used.)
Then with some effort one can show $\mathcal{L}$ is Lindbladian if and only if $$\qquad\qquad\qquad\qquad\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P} \ge 0,\qquad\qquad\qquad\qquad (1) $$ where $\mathcal{P} \equiv \mathcal{I} - \mathcal{I}^{\mathrm{PT}}/N = \mathcal{P}^2$ is the "superprojector" that removes an operator's trace, so that $\mathcal{P}[B] = B - (\mathrm{Tr}[B]/N)I$. Note that $\mathcal{S} \ge 0$ means that a superoperator $\mathcal{S}$ is a positive operator (when considered as an operator on the space of operators/matrices) in the sense of being Hermitian with positive eigenvalues or, equivalently, that $\langle B, \mathcal{S}[B]\rangle \ge 0$ for all operators $B$, where $\langle B, C \rangle \equiv \mathrm{Tr}[B^\dagger C]$ is the Hilbert-Schmidt inner product on the space of operators. This is a distinct condition from $\mathcal{S}$ being positivity preserving, i.e., $B\ge 0 \Rightarrow \mathcal{S}[B]\ge 0$, which is (confusingly) usually described as $\mathcal{S}$ being a "positive map".
An equivalent condition to Eq. (1) is $$\qquad\qquad\overline{P}_\Psi[ (\mathcal{L}\otimes \mathcal{I})(|\Psi \rangle\langle \Psi|)] \overline{P}_\Psi \ge 0,\qquad\qquad(2)$$ where $|\Psi \rangle = N^{-1} \sum_{n=1}^N|n\rangle|n\rangle$ is some maximally entangled state and $\overline{P}_\Psi=I - |\Psi \rangle\langle \Psi|$ projects onto the orthogonal subspace. (This condition is independent of the choice of basis $\{|n\rangle\}$ and hence the choice of maximally entangled state $|\Psi \rangle$.) Eq. (2) is also a condition about positivity of a linear operator, but in this case it's a condition on a tensor product of two ($N \times N$) density matrices rather than a condition on a single ($N^2 \times N^2$) superoperator as in Eq. (1).
Eq. (2) is the form the Lindbladian condition appears in some monographs like Wolf's "Quantum Channels and Operations: Guided Tour" [PDF] (see eq. (7.15)) and, I think, Tarasov's "Quantum Mechanics of Non-Hamiltonian and Dissipative Systems" (see Sec. 15.8 and 15.9). I prove Eq. (1) in a self-contained and elementary way in a blog post here.
If we want to further check whether $\mathcal{L}$ is completely dissipative and hence generates trace-preserving evolution (for all $B$, $\mathrm{Tr}[e^{t\mathcal{L}}[B]] = \mathrm{Tr}[B]$ or, equivalently, $\mathrm{Tr}[ \mathcal{L}[B]]=0$), then we just need to confirm a vanishing partial-trace condition, $$0 = \sum_{p=1}^N \mathcal{L}_{(pp)(nm)},$$ using the index convention $(\mathcal{S}[B])_{nn'} = \sum_{m,m'=1}^N \mathcal{S}_{(nn')(mm')}B_{mm'}$.