Supposing we have a sequence of numbers $a_n$ that ranges from 1 to a particular hypernatural $\delta$. This is an infinite series in the sense that it "goes past" the natural numbers and the natural numbers are countably infinite. However, this sequence does have an end, in this case $\delta$.
The cardinality of the natural numbers is $\aleph_0$ and the cardinality of the hypernaturals is the same as the real numbers, $2^{\aleph_0}$.
My question is what is the cardinality of $a_n$? Is it $2^{\aleph_0}$, $\aleph_0$, or something in-between the two?
Guessing, I would say that it is in-between the two as it goes past the cardinal number $\aleph_0$ but I am not sure if this reasoning even works.
Cardinality is something you want to assign to sets, but a sequence is a function from $\mathbb N$ (or in this case $^*\mathbb N$) onto another set. So your question seems to not make sense.
You could of course ask what the cardinality of $\{1,\ldots,\delta\}$ is. (by the way why do you chose $\delta$ as a hypernatural, this symbol is usually reserved for infinitesimal entities)
To which the answer would be it's either $\aleph_0$, if $\delta$ is finite, or $\aleph_1$ if $\delta$ is unlimited (i.e. $\delta\in{}^*\mathbb N\setminus\mathbb N$), if we assume that the continuum hypothesis holds true.
Edit: Actually we can prove that in the latter case it is $\aleph_1$ regardless of the continuum hypothesis. This is due to the structure of the hypernaturals (I'm assuming we use the standard ultrafilter construction of ${}^*\mathbb N$)
Claim: $\forall \delta\in {}^*\mathbb N\setminus\mathbb N\;\forall r\in\mathbb R_+\exists q\in {}^*\mathbb Q_+ : q\delta \in \text{hal}(r\delta)\cap{}^*\mathbb N$
Proof: Let $\mathcal F$ be our ultra-filter. Then $\delta$ is associated with a sequence $d_{n} \in \mathbb R^{\mathbb N}$ such that $\{n\in\mathbb N : d_n \in\mathbb N\}\in \mathcal F$. Since $\delta$ is assumed to be unlimited, this sequence must go to infinity. Now we simply take a sequence $b_n\in\mathbb N^{\mathbb N}$ such that $\lim\limits_{n\to\infty}|r-\frac{b_n}{d_n}| = 0$. Consequently let $q_n = \frac{b_n}{d_n}$ and $q\in{}^*\mathbb Q$ be the hyperrational number associated with that sequence. Then obviously $q\in\text{hal}(r)$, hence also $^{(1)}$ $q\delta\in\text{hal}(r\delta)$. But on the other hand $q\delta$ corresponds to the sequence $\frac{b_n}{d_n}d_n = b_n$, hence $q\in{}^*\mathbb N$.
Remark: So in fact if you have any unlimited hypernatural $\delta$ and take any positive real number $r$ then there is another hypernatural $\delta'$ such that $|r\delta-\delta'|$ is infinitesimal.
(1): Here we have to be a bit more careful, e.g. if $q = r +\epsilon$, $\epsilon$ infinitesimal, then $q\in\text{hal}(r)$, but $\frac{1}{\epsilon}q \notin \text{hal}(\frac{1}{\epsilon}r)$. We can fix this issue in our proof by taking $q$ close enough to $r$, i.e. such that $\delta|q-r|$ is still infinitesimal.