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 $\lim_{n \rightarrow \infty} \operatorname{sup}_{x \in M} \sum_{k=n}^{\infty} \left\lvert x_k \right\rvert=0$ 

EDIT: Since I received an answer that outlined to me that it does not work in the non-symmetric case, I thought it would be good to explain why I think that symmetry could help. In this case, I can show that $\sigma(T) \subset \sigma(T_{\vert_{\ell^1}})$ and the point spectra coincide. Moreover, also all finite-dimensional eigenspaces are the same for each element of the point spectrum. In other words, the spectrum of $T\vert_{ \ell^1}$ also contains the spectrum that one would assume to have for a compact operator. However, I could so far not exclude any continuous spectrum for $T\vert_{ \ell^1}$ which would be a necessary condition for compactness.