Suppose $q:X \rightarrow Y$ is a quotient map of topological spaces such that the product map $q^2:X^2\rightarrow Y^2$ is also a quotient map. Are the maps $q^n:X^n\rightarrow Y^n$ quotient maps for all $n\geq 3$? If not, are there sufficient conditions that make this the case?
If you add the hypothesis that $q$ is an open map, then your conclusion follows for all powers, including infinite powers, and there is no need in this case separately to assume that $q^2$ is a quotient map.
Theorem. If $q:X\to Y$ is a continuous surjective open map, then all powers $q^I:X^I\to Y^I$ are quotient maps.
Proof. The Wikipedia page on quotient maps mentions that it is a sufficient condition (but regrettably not necessary) for a map to be a quotient map that it is continuous, surjective and open. Suppose that $q:X\to Y$ is continuous, surjective and open. It follows using standard methods that the power map $q^I:X^I\to Y^I$ is also continuous, surjective and open. So it is a quotient map.QED
If you drop the open map hypothesis, however, then I'm not sure what happens. (I couldn't manage to make Herb's suggestion work in the converse direction.) Perhaps one can build a crazy counterexample with a non-open quotient map.