Let $T$ be a compact symmetric operator on $\ell^2$ and $T\vert_{\ell^1}$ be bounded on $\ell^1$. Are there any non-trivial conditions that $T\vert_{\ell^1}$ is compact as well (for example would $T$ belonging to some Schatten-class on $\ell^2$ be sufficient)?

The obvious proof estimating

$$\left\lVert \sum_{i=0}^{\infty} \lambda_i \langle \cdot ,\varphi_n \rangle \varphi_n - \sum_{i=0}^{k} \lambda_i \langle \cdot ,\varphi_n \rangle \varphi_n \right\rVert_{L(\ell^1)} $$ does not work as the eigenvectors, we get from the $\ell^2$ representation, are not necessarily bounded in $\ell^1$ norm.

Recall also that a set $M \subset \ell^1$ is compact if it is bounded, closed and $\lim_{n \rightarrow \infty} \operatorname{sup}_{x \in M} \sum_{k=n}^{\infty} \left\lvert x_k \right\rvert=0$