If by $\{ \aleph_0, \aleph_1, \aleph_2, \dots \}$ you mean $\{ \aleph_n : n \in \mathbb{N} \}$, then this set is countable because it's indexed by the natural numbers. However, this set doesn't contain $\aleph_{\omega}$ or any larger alephs.
If by $\{ \aleph_0, \aleph_1, \aleph_2, \dots \}$ you mean 'the set[sic] of all (well-orderable) cardinals', then it's certainly not countable; in fact, it's too big to be a set! (Hence the 'sic'.) There is an aleph for each ordinal $\alpha$, and there are class-many ordinals.
It should also be noted that, unless you accept the generalised continuum hypothesis, it is not necessarily the case that $\aleph_{\alpha} = |\mathscr{P}^{\alpha}(\mathbb{N})|$ for all $\alpha$, as you suggest in your question.
Ordinals are not cardinals.
Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.
So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.
Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.
But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $\aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $\aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.
This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.
Okay, what's the point of all that?
The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".
When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.
So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $\omega$ is the appropriate notation, since it denotes an ordinal, rather than $\aleph_0$ which denotes a cardinal.
Between $\aleph_\omega$ and $\aleph_{\omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".
Note that $\aleph$ numbers are not defined by power sets, these are $\beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.
Best Answer
Much like I wrote in Cardinal number subtraction, if $\kappa$ and $\lambda$ are two $\aleph$-numbers, it might be possible to define division, but this definition would have to be limited and awkward.
If $\kappa$ and $\lambda$ are both regular cardinals and $\kappa<\lambda$ then every partition of $\lambda$ into $\kappa$ many parts would have to have at least one part would be of size $\lambda$. In a sense this means that $\frac\lambda\kappa=\lambda$. This is indeed the case with $\aleph_1/\aleph_0$, both are regular cardinals are $\aleph_0<\aleph_1$.
If, however, $\kappa=\lambda$ this is no longer defined, since $\kappa=2\cdot\kappa=\aleph_0\cdot\kappa=\ldots=\kappa\cdot\kappa=\ldots$, so there can be many partitions of $\lambda$ into $\kappa$ many parts, and in each the parts would vary in size (singletons; pairs; countably infinite sets; etc.)
When $\lambda$ is a singular limit cardinal, e.g. $\aleph_\omega$ this breaks down completely, since singular cardinals can be partitioned into a "few" "small" parts. In the $\aleph_\omega$ case these would be parts of size $\aleph_n$ for every $n$, which make a countable partition in which all parts are smaller than $\aleph_\omega$.
The only reasonable way I can think that cardinal division can be defined would have to consider the Surreal numbers, and the embedding of the ordinals in them. However this will not be compatible with cardinal arithmetic at all (the surreal numbers form a field).
I should also remark that your reasoning for $\aleph_1/\aleph_0$ being $\aleph_1$ is invalid. First note that neither is a real number, and that it is possible that $\aleph_1$ is much smaller than the cardinality of the real numbers (so between two natural numbers there are a lot more real numbers). Secondly, note that between two rational numbers there are also infinitely many rational numbers - does that mean $\aleph_0/\aleph_0=\aleph_0$?
However your rationale is not that far off, as I remarked in the top part of the post, if you take a set of size $\aleph_1$ and partition it into $\aleph_0$ many parts you are guaranteed that at least one of the parts would have size $\aleph_1$.
Further reading: