Given a index set $S$ together with a ultrafilter $\mu$ on $S$ (such that no set of cardinality $< |S|$ has measure $1$). Let the ordered field $\mathbb{R}(S,\mu)$ denote the ultrapower of $\mathbb{R}$ with respect to $S$, i.e.

$\mathbb{R}(S,\mu):=\mathbb{R}^S/\sim$, where two maps $f,g$ are called equivalent, if $\mu(\{i\in S| f(i)=g(i)\})=1$. This is again an ordered field. Equip it with the topology generated by

$\{B_\varepsilon(x)|x\in \mathbb{R}(S,\mu),\varepsilon \in \mathbb{R}(S,\mu),\varepsilon>0\}$.

Then my question is: What is the smallest cardinality of a local basis around $0$ depending on the cardinality of |S| and (possibly) on the ultrafilter?

Examples:

If $S=\{pt\}$, we get $\mathbb{R}(S,\mu)\cong \mathbb{R}$ and hence it has a countable base for the topology. In the case $S=\mathbb{N}$ and a non-principal ultrafilter one can show, that there is no countable base for the topology (saturation argument) and it is at most $|\mathbb{R}(\mathbb{N},\mu)|=|\mathbb{R}|$. CH would tell us, that it is $|\mathbb{R}|$. But maybe there is a good (avoiding CH) reason, why it has exactly this cardinality.