When does proving Tychonoff's theorem not require extensions of the ZF set theory? Proving Tychonoff's theorem requires the axiom of choice, in general. For Hausdorff spaces, just the Boolean prime theorem is enough. Is it possible to identify interesting classes of spaces or of index sets for which just the ZF framework is enough?
In particular, what happens (if anything noteworthy) for countably many Riemannian manifolds?
Of course, I am not interested in trivial answers (such as a finite number of non-empty spaces).
 A: As an overview of how complicated the relation Tychonoff / AC is, see these papers which show that for other alternative definitions of compactness (which might be non-equivalent in the absence of AC) we see that the corresponding version of Tychonoff might or might not be equivalent to AC (or weaker versions). (See this paper for a comparison of the definitions themselves).
You don't need Tychonoff for a large class of spaces to prove the axiom of choice: the proof here shows that if we assume

The Tychonoff theorem holds for the class of all topological spaces $(X,\mathcal{T})$ such that $|\mathcal{T}| =4$ (which are obviously all compact)

Then full AC must already be true.
In ZF we can certainly show it for the class of all spaces with $|\mathcal{T}|= 2$  because then $X$ is indiscrete and any product of indiscrete spaces is indiscrete hence compact.
Maybe AC is already implied by Tychonoff for topologies of size 3? I think so, because it suffices (in the linked proof) to define the topology as $\{X_i, \{p_i\}, \emptyset\}$, I think, as @bof also noticed. 
As to your countable case, this is much weaker. It might even be in ZF if your subclass of spaces (like Riemannian manifolds) is small enough. I'll let others be the judge of that. E.g. it might make a difference of the spaces are second countable and whether you want sequential compactness or other forms)
In summary: it's subtle.
