You have that $T^*:(\ell^2)^* \rightarrow E^{**}$, so using that $(\ell^2)^*$ is isomorphic to $\ell^2$ (just in the linear sense, as we already have a co-ordinate system), we can regard $T^*$ as a map $\ell^2\rightarrow E^{**}$.

By the Principle of Local Reflexivity (I've used a paper of Behrends in the past, which is overkill, but is freely available: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10022.pdf Or look in a book on Banach space theory) for each triple $i=(M,N,\epsilon)$, where $M\subseteq E^{**}$ and $N\subseteq E^*$ are finite-dimensional, and $\epsilon>0$, we can find an operator $S_i:M\rightarrow E$ such that $(1-\epsilon)\|x\| \leq \|S_i(x)\| \leq (1+\epsilon)\|x\|$ for $x\in M$, and with $\phi(S_i(x)) = x(\phi)$ for $x\in M$ and $\phi\in N$.

So, let $P_n:\ell^2 \rightarrow \ell^2$ be the projection onto the first $n$ co-ords, let $M\supseteq T^*(P_n(\ell^2))$ and let $i=(M,N,\epsilon)$, so $S=S_i T^* P_n$ makes sense, and is a map $\ell^2\rightarrow E$. Then, for $a\in\ell^2$ and $\phi\in E^*$, $$S^*(\phi)(a) = \phi(S_i T^* P_n(a)) \rightarrow T^*(a)(\phi) = T(\phi)(a),$$ as $i$ and $n$ increase.

So we have found a bounded net $(S_d)$ (we can even choose it bounded by $\|T\|$ be rescaling a little) with $S_d^* \rightarrow T$ in the weak operator topology. But now a standard trick (take convex combinations, as the closure of a convex set is the same in the weak and norm topologies) allows us to find a net which converges SOT.

I think the proof would work for any Banach space F replacing $\ell^2$, as long as we can find a bounded net of finite rank operators $(F_\alpha)$ with $F_\alpha\rightarrow 1$ SOT. That is, F should have the bounded approximation property.