The enclosed characterization is not completely satisfactory, but
it is far from being clear that it is possible to get a better
one. Observe that since $c_0$ is a closed subspace of
$\ell_\infty$, the closure is in $L(\ell_2,c_0)$. It does not
exhaust all of $L(\ell_2,c_0)$ as is shown in the answer to:
https://mathoverflow.net/questions/184473/

Now let $B$ be an operator in the closure. It is standard that
this happens if and only if we can find a sequence $B_i\in
L(\ell_2,\ell_2)$ such that $||B_i||_{2\to\infty}\le 2^{-i}$ for
$i\ge 2$ and $B=\sum_{i=1}^\infty B_i$ (convergence in
$L(\ell_2,\ell_\infty)$). This implies that the sequence
$\{Be_j\}$, where $\{e_j\}$ is the unit vector basis of $\ell_2$
can be decomposed as $Be_j=\sum_{i=1}^\infty B_ie_j$ (in
$\ell_\infty$). Now let us consider sequences
$\{B_ie_j\}_{j=1}^\infty$. This sequence for $i\ge 2$ has two
properties: (1) As sequence in $\ell_2$ it is majorated by the
orthonormal basis (in the sense that it is the image of the
orthonormal basis under a bounded linear operator
$\ell_2\to\ell_2$) and (2) $||\sum_{j=1}^\infty
\alpha_jB_ie_j||_\infty\le 2^{-i}$ for every
$\{\alpha_j\}_{j=1}^\infty\in\ell_2$.

The argument can be reversed. Of course, this condition is not
handy, but possibly it can be used for some purposes. For example, it can be used to disprove some of the possible conjectures about the closure.