(Warning: The OP asked for "symmetric" (aka "self-adjoint") and this example is not.)
Here's an example showing that $T$ can be trace-class but $T|_{\ell^1}$ is not compact.
Let $(x_n)$ be a sequence of vectors in $\ell^2$ with disjoint supports, $\sum_n \|x_n\|_2 \leq 1$ and $\|x_n\|_1=1$ for all $n$. Define $$ T(\xi) = \sum_n \xi_n x_n \qquad (\xi\in\ell^2). $$ Then $\| T(\xi) \|_1 \leq \sum_n |\xi_n| \|x_n\|_1 \leq \|\xi\|_1$ so $T$ is bounded on $\ell^1$. However, $(T(e_n)) = (x_n)$ has no convergent subsequence so $T$ is not compact on $\ell^1$. As $\sum_n \|x_n\|_2\leq 1$, $T$ is trace-class.
An example of such a sequence is as follows. Let $N(n)$ be a rapidly increasing sequence of integers, and choose $x_n$ to be the sequence $(0,\cdots,0,N(n)^{-1},\cdots,N(n)^{-1},0,\cdots)$ where we repeat $N(n)^{-1}$ exactly $N(n)$ times, and we place the non-zero terms so that the $x_n$ have disjoint support. Then $\|x_n\|_1=1$ but $\|x_n\|_2 = N(n)^{-1/2}$ so as long as $N(n)$ increases fast enough that $\sum_n N(n)^{-1/2} \leq 1$ we're done.
This $T$ is not self-adjoint, but notice that $T^*$ is trace-class, and $T^*$ is still bounded on $\ell^1$. Furthermore, $S=T+T^*$ is seen to still be such that $S(e_n)$ has no convergent subsequence in $\ell^1$.