The explicit questions
"Is zero the limit to the following sequence?"
Yes. Unless you clarify that you're doing something very nonstandard, you must be using one of the compatible default meanings of limit everyone would expect, with a context of rationals or real numbers, and the limit is indeed $0$.
Why it shouldn't be [something other than $0$]?
There is good reason for this: in the context of the reals or the rationals, $0$ is the unique number $\ell$ for which the differences $1/2-\ell, 1/4-\ell,\ldots$ become arbitrarily small (smaller than any positive rational/real). The reals/rationals don't have any number greater than all of the $2^n$s (like "$\aleph_0$") to let you build a counterexample by taking the reciprocal.
In a sense, the real numbers are the maximal ordered field without "infinite"/"unbounded" elements greater than all of the integers (this is the archimedean property).
$\aleph_0$ is the limit to sequence $2,4,8,\dots$
This is incorrect. Under the usual definition of limit, we would usually say this diverges (since it has no real number limit) to $\infty$ (see wikipedia on this), and the $\infty$ symbol doesn't really have a connection to any cardinality. In some contexts (e.g. topology), you could think of $\infty$ as a point in a special "real line with endpoints" called the extended reals. But even then, there wouldn't be a good reason to associate $\infty$ with any other cardinal.
Additionally, the claim is not easy to fix even with a charitable reading. Even if you wanted to let some sort of infinite number be some sort of limit of the sequence $2,4,8,\dots$, it would be reasonably be a number where you could add $1$ and get a new number. Using cardinal notation like $\aleph_0$ strongly implies you want something where $\aleph_0+1=\aleph_0$ (see cardinal addition), which isn't likely to work as a nice infinite number you would use here.
In a comment, you mentioned the fact that in standard set theory we take $\aleph_0$ to be $\omega_0$ (a.k.a. $\omega$), and claimed that this doesn't make a difference. But when you're trying to explain a non-standard idea like this, even if there's technically no difference in denotation, the difference in connotation (in what it implies about the expected properties of addition) is significant.
Is there something close to the intuition?
As Arthur suggested in a comment, in non-archimedean ordered fields and similar (like the surreal numbers), there is some hope since there are infinite and infinitesimal elements.
Indeed, in the surreal numbers, there is a copy of the ordinals, except that the arithmetic operations aren't the usual ones, they're the so-called natural sum and product. Notably, $\omega+1$ works out the same either way.
To talk about examples like the one in the question, we don't need all of the surreals, just a chunk that has numbers like $\omega$ in it.
"Limits" for the surreals
In several works of Norman L. Alling on the Surreals, including the paper Conway's Field of Surreal Numbers, Alling has defined a sense of "limit" different than the usual one, for which certain sequences (and similar - the indexing set could be any limit ordinal) have unique surreal limits.
Very briefly, if you have a sequence that is "very Cauchy" (my phrasing) in the sense that the difference between $|a_k-a_j|$ is infinitesimal compared to $|a_j-a_i|$ whenever $i<j<k$, then there are many "pseudo-limits", but there is a unique pseudo-limit that is the simplest (in the sense of simplicity/birthday of the surreals), which Alling calls "the limit".
This definition, while interesting, does not apply to $1/2,1/4,1/8,\ldots$ as that sequence is merely Cauchy in the usual sense, not "very Cauchy".
"Limits" for the surreals - take 2
We can take still take inspiration from Alling's work, though. We can just take the work to find pseudo-limits and hope that it'll still make some sense. (In doing so, we replace his original notion of pseudo-limits with one that uses absolute value in place of a valuation of the location in a hierarchy of infinitesimals.)
For example, $1/3,1/9,1/27,\ldots$ would have to have something like pseudo-limits between lower bounds $1/9-|1/3-1/9|=-1/9, 1/27-|1/9-1/27|=-1/27,\ldots$ and upper bounds $1/9+|1/3-1/9|=1/3, 1/27+|1/9-1/27|=1/9,\ldots$. There are many surreal numbers between those negative and positive bounds, but the simplest one is $0$.
But if we take the example from the question, $1/2,1/4,1/8,\ldots$ would have to have something like pseudo-limits between lower bounds $1/4-|1/2-1/4|=0, 1/8-|1/4-1/8|=0,\ldots$ and upper bounds $1/4+|1/2-1/4|=1/2, 1/8+|1/4-1/8|=1/4,\ldots$. There are many surreal numbers between $0$ and the $1/2^n$s, but the simplest one is $1/\omega$.
We still may not say that "the limit of $1/2,1/4,1/8,\ldots$ is [anything but $0$]". However, we could say something like "If you extend the definitions of Alling for surreal limits to certain cauchy sequences where something reminiscent of pseudo-limits still happen to exist, then the 'limit' of $1/3,1/9,1/27,\ldots$ is $0$, and the 'limit' of $1/2,1/4,1/8,\ldots$ is $1/\omega$."
A note about the reciprocals
Note that even with my implied re-definition of Alling's definition of surreal limits, sequences like $2,4,8,\ldots$ still don't have limits because they're not anything like Cauchy. Taking inspiration from the monotone convergence theorem, since it's increasing, we might want to say that its "limit" should be the simplest (in the sense of the surreals) upper bound.
In that sense, $2,4,8,\ldots$ and $3,9,27,\ldots$ both have "limit" $\omega$. This is unfortunate for the intuition in the question, since the sequences of reciprocals have (for a different meaning of "limit") two different limits: $0$ and $1/\omega$.
As an aside, since $1/2,1/4,\ldots$ and $1/3,1/9,\ldots$ are decreasing, we might want to examine the simplest lower bound for each sequence. Fortunately or unfortunately, the answer is $0$ for both.