First of all, the axiom of countable choice says that given a countable family of non-empty sets, you can choose from each set *simultaneously*. If you want to choose from one, then from another, then from another, and so on you need a strictly stronger form of choice called **Dependent Choice**, abbreviated as $\sf DC$.

To your questions, the definition of the $\aleph$ numbers uses absolutely no choice, although the proof that any of them is regular does use the axiom of choice (well, except $\aleph_0$). So it is consistent, for example, that $\aleph_1$ is singular. But assuming $\sf DC$ will prevent that. Whether or not you can have every cardinal $\geq\aleph_2$ singular with $\sf ZF+DC$ is still open.

The definition of $\beth$ numbers, on the other hand, goes out the window. The axiom of choice is equivalent to saying that the power set of a well-ordered set is well-ordered. So if the axiom of choice fails, there will be some ordinal $\alpha$ whose power set cannot be well-ordered. This means that at some point $\beth$ cardinals will not be $\aleph$ numbers. You can still talk about $\beth$ numbers, of course, as iterated powers of $\omega$, but that gives you significantly less information in that sense.

As for the continuum hypothesis, as noted by others it is still unprovable. But now you even get different forms of the continuum hypothesis. $2^{\aleph_0}=\aleph_1$ is no longer equivalent to "Every uncountable set of reals is equipollent with the reals themselves". Indeed, even without $\sf DC$, it is consistent that $2^{\aleph_0}$ and $\aleph_1$ are incomparable, and every uncountable set of reals has cardinality $2^{\aleph_0}$.

Finally, for large cardinals, most properties which are reserved for very large cardinals can be made compatible with $\aleph_1$ and $\sf ZF+DC$. For a more complete survey, look at What sort of large cardinal can $\aleph_1$ be without the axiom of choice?.

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