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: Is $L(\ell_2,\ell_2)$ dense in $L(\ell_2,c_0)$?Is $L(\ell_2,\ell_2)$ dense in $L(\ell_2,c_0)$?
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.
