Let $|I|\leq \aleph _\alpha$, and let $ \left\{ \beta _i \right\} _{i\in I}$ be cardinals with $\beta _i \leq \aleph _{\alpha}$.
Is it true that $\aleph_{\alpha +1} > \sum _{i\in I}\beta _i $?
Using choice, I think it's possible to write $\aleph _{\alpha+1}=\sum _{i\in I}\aleph_{\alpha +1}$, and I think $$\sum _{i\in I}\aleph_{\alpha +1} > \sum _{i\in I}\beta _i $$ holds because $|I|\leq \aleph _\alpha$, but I'm not sure whether this is true nor how to prove it.
You're on the right track about the inequality $$\sum_{i\in I}\aleph_{\alpha+1}>\sum_{i\in I}\beta_i.$$ We know each $\beta_i$ is $\le \aleph_\alpha$, in addition to $\vert I\vert\le\aleph_\alpha$, so we have $$\sum_{i\in I}\beta_i\le \sum_{i\in I}\aleph_\alpha\le\aleph_\alpha\times\aleph_\alpha=\aleph_\alpha.$$ Note that we aren't using choice anywhere, here! (The statement "for all infinite $X$, $\vert X\times X\vert=\vert X\vert$" is indeed equivalent to choice, but its restriction to well-orderable $X$s - that is, to $\aleph$s - is just provable in ZF.)
Now, a side note: there is a surprising subtlety here! If we drop the axiom of choice, we can have $\aleph_1$ be a union of countably many countable sets. But I wrote that we didn't need choice in the argument above! So, what gives?
The answer is the crucial inequality $$\sum_{i\in I}\beta_i\le\sum_{i\in I}\aleph_\alpha.$$ This requires us to have a set of maps $f_i: \beta_i\rightarrow \aleph_\alpha$. In case each $\beta_i$ is actually a cardinal, then there is a canonical map from each $\beta_i$ to $\aleph_\alpha$: the initial segment map! This doesn't require choice.
But now what if we just imagine we have a collection of sets $B_i$ ($i\in I$) and each $B_i$, individually, has cardinality $\le\aleph_\alpha$. Then this means that for each $i$, there is some $g_i$ which is an injection from $B_i$ into $\aleph_\alpha$. However, without the axiom of choice, we can't pick out a specific such $g_i$ for each $i\in I$ "all at once."
To put it in other words: there are some countable sets which have no "nice" bijection with $\aleph_0$. That is, they do have bijections with $\aleph_0$ - they're countable, after all - but none of them are nicely definable. For example: given a countable ordinal $\delta$, there is in general no "best" way to biject $\delta$ with $\aleph_0$. This is essentially the key to understanding how $\aleph_1$ can be a countable union of countable sets.