Implications of continuum hypothesis and consistency of ZFC
Solution 1:
You mixed the quantifiers. Assuming the consistency of $ZFC$, it is consistent that for every $n>0$ there is a model of $ZFC$ such that $\frak c=\aleph_n$.
In fact, the result is even better. Let us introduce a new term:
We say that an ordinal $\alpha$ has cofinality $\omega$ ($\omega=\aleph_0$, the set of natural numbers) if we can find an increasing sequence $\langle\alpha_n\mid n<\omega\rangle$ such that $\alpha_n<\alpha$ and $\sup_n\alpha_n=\alpha$.
For example, $\omega_1+\omega$ has cofinality $\omega$, simply by $\alpha_n=\omega_1+n$. However $\omega_1$ does not have cofinality $\omega$ since a countable union of countably ordinals is countable.
Theorem: Let $\alpha$ be a finite number or an ordinal whose cofinality is not $\omega$, it is consistent that $\frak c=\aleph_\alpha$.
(Such result is proved through forcing. I will not get into the proof.)
Using another theorem we also have that this is pretty much all there is to say about this problem.
Theorem: The continuum does not have cofinality $\omega$.
This is of course a minor corollary from a much more general case, however it gives us that if $\alpha$ is a finite number, or does not have cofinality $\omega$ then it is possible that $\frak c = \aleph_\alpha$.
So to your original question: For all $n$ it is consistent with $ZFC$ that $\frak c=\aleph_n$. It does not mean that it is consistent with $ZFC$ that for all $n$, $\frak c=\aleph_n$.
From the above theorems, we have that $\aleph_\omega$, the first cardinal which is bigger than all the $\aleph_n$ cannot be $\frak c$. So not every uncountable cardinal can have the cardinality of the continuum.
Solution 2:
The answer to your question is no, here's why. Even if it's true for each $n \in \mathbb N$ the theory $\textbf{ZFC}+\mathfrak c = \aleph_n$ is consistent, it's not true that for any pair $n,m \in \mathbb N$ the theory $\mathbf{ZFC}+\mathfrak c = \aleph_n+\mathfrak c = \aleph_m$ is consistent, indeed this is consistent if and only if $n=m$. The reason for that is that in ZFC you can prove that for each pair $n,m \in \mathbb N$ you have $\aleph_n=\aleph_m$ just when $n=m$, if add those two axiom above to ZFC with $n \ne m$ then you get an inconsistent theory.
Hope this could help you.