How many epsilon numbers $<\omega_1$ are there?

Note that $\varepsilon_0$ is countable. So if there are only countably many countable $\varepsilon$ numbers, they would have a countable supremum, $\alpha$. Consider now the same construction as $\varepsilon_0$, starting $\alpha+1$. That is: $$\sup\{\omega^{\alpha+1},\omega^{\omega^{\alpha+1}},\ldots\}$$

The result is itself an $\varepsilon$ number, and it is countable (as the countable limit of countable ordinals). But all the countable $\varepsilon$ numbers were assumed to be below $\alpha$, which is a contradiction.

Your intuition is correct: $\omega_1=\epsilon_{\omega_1}$. See the discussion in Wikipedia: the constructions of $\epsilon_{\alpha+1}$ from $\epsilon_\alpha$ and of $\epsilon_\alpha$ for a countable limit ordinal $\alpha$ preserve countability, so there must be $\omega_1$ countable $\epsilon$-numbers.