The answer is no. The continuum cannot be $\aleph_\omega$, and this can be proved in ZF, that is, without using the axiom of choice. To see
this, suppose towards contradiction that $P(\omega)$ is equinumerous with
$\aleph_\omega$. Since $P(\omega)$ is equinumerous
with $P(\omega)^\omega$, and this does not require AC, it follows
that there is a bijection $f:\aleph_\omega\cong
(\aleph_\omega)^\omega$. Let $g(n)$ be the first ordinal
$\alpha\lt\aleph_\omega$ that is not among $f(\beta)(n)$
for any $\beta\lt\aleph_n$. Since there are fewer than
$\aleph_\omega$ many such $\beta$, it follows that there
are fewer than $\aleph_\omega$ many such $f(\beta)(n)$, and
so such an $\alpha$ exists. Thus, $g:\omega\to
\aleph_\omega$. But notice that for any particular
$\alpha\lt\aleph_\omega$, we have $\alpha\lt\aleph_n$ for
some $n$ and consequently $g(n)\neq f(\alpha)(n)$, and thus
$g\neq f(\alpha)$. Thus, $f$ was not surjective to
$(\aleph_\omega)^\omega$ after all, a contradiction.

This is just a standard proof of Konig's theorem (that
$\aleph_\omega^\omega\gt\aleph_\omega$), and the point is
that it doesn't use AC.