The splitting number $\mathfrak{s}$ is the minimal cardinal $\kappa$ such that if $\mathcal{A} \subseteq \mathcal{P}(\omega)$ has size $< \kappa$, there exists some $X \subseteq \omega$ such that for all $A \in \mathcal{A}$, $X \setminus A$ is finite or $X \cap A$ is finite.

Then if all $X_i$ are sequentially compact Tychonoff spaces, and $|I| <\mathfrak{s}$ then $\prod_{i \in I} X_i$ is also sequentially compact. And $\{0,1\}^{\mathfrak{s}}$ is not sequentially compact. What $\aleph_\alpha$ equals $\mathfrak{s}$ cannot be said in ZFC. It could be $\alpha = 1$ (under CH, e.g.) or much bigger, depending on the size of the continuum. See van Douwen's paper in the Handbook of Set Theoretic Topology on cardinal invariants of the continuum.

not metrizable, holds for $\beta =\aleph_1$, but not for any cardinal strictly less. I doubt this is what you had in mind. $\endgroup$ – David Handelman Nov 11 '15 at 3:26