From what I understand, it's independent of ZFC whether $2^{\aleph_0} > \aleph_\alpha$ for any nonzero ordinal $\alpha$. But there are larger cardinalities than the $\aleph$ numbers, especially if we start including large cardinal axioms. So my question is twofold.
- Can ZFC alone prove the existence of a cardinal $\kappa$ such that $2^{\aleph_0} \le \kappa$?
- Do any of the large cardinal axioms together with ZFC imply the existence of a $\kappa$ with $2^{\aleph_0} \le \kappa$?
The axiom of choice is equivalent to stating that every infinite cardinal is an $\aleph$ number. We can talk about non-$\aleph$ cardinals in the absence of choice, but it's still true there is a proper class of $\aleph$ cardinals, regardless of assuming choice or not.
The continuum, therefore, is certainly an $\aleph$ number under $\sf ZFC$, and since we cannot prove that $2^{\aleph_0}=\aleph_1$ or not, there is no ordinal that $\sf ZFC$ proves is the $\aleph_\alpha$ of the continuum. What $\sf ZFC$ does prove is that if $2^{\aleph_0}=\aleph_\alpha$, then $\alpha>0$ and $\operatorname{cf}(\alpha)>\aleph_0$; and we know that this is the only provable restriction on the size of the continuum from the works of Cohen and Solovay.
To recap: assuming $\sf ZFC$, the continuum is an $\aleph$ cardinal.
Large cardinal axioms do postulate the existence of "large cardinals", but this is a technical term—without a well-defined meaning—and not a term for a cardinal larger than all the $\aleph$'s. It is just usually the case that if $\kappa$ is a large cardinal, then $\kappa=\aleph_\kappa$ (not always, though!) so it is sort of hard to describe $\kappa$ using the $\aleph$ numbers.
Moreover, due to the work of Levy and Solovay, we also know that most large cardinal axioms are compatible with both the continuum hypothesis and its negation, so large cardinals do not determine the value of the continuum either.