NOTE: The following proof is valid if one defines an “operator” as a linear function, not necessarily assumed to be bounded, between two vector spaces.
I claim that if $X$ is as you described, then it must be finite-dimensional.
To see this, suppose, for the sake of contradiction, that $X$ is infinite-dimensional. Then, the dimension of $X$ must be at least of the cardinality of the continuum. (See here. In fact, one can use Baire’s category theorem to conclude that the dimension of $X$ must be uncountable, but the former result is stronger without assuming the continuum hypothesis. Why one needs the stronger result that $\operatorname{dim} X\geq\#\mathbb R$ will be clear below.) The dimension of $\ell^p$, on the other hand, is precisely $\#\mathbb R$ for any $p\in(1,\infty)$. (After all, the cardinality of all real sequences is $\#\mathbb R$.) Let’s take a Hamel basis $\mathscr H$ of $X$ and a Hamel basis $\mathscr L$ of $\ell^p$.
By the preceding arguments, there exists a surjective function $f:\mathscr H\to\mathscr L$. Using the fact that basic representations are unique, one can extend $f$ to a surjective linear function $F:X\to\ell^p$. By assumption, $F$ is a compact operator, so it is a fortiori continuous. Letting \begin{align*} U\equiv\{x\in X\,|\,\|x\|<1\},\\ C\equiv\{x\in X\,|\,\|x\|\leq 1\}, \end{align*} one has that $F(C)$ is precompact in $\ell^p$, so that $F(U)$ is also precompact. Invoking the open-mapping theorem, one can conclude that $F(U)$ is a non-empty, precompact open set. But this is impossible, since $\ell^p$, being an infinite-dimensional normed vector space, cannot be locally compact.