As explained in my answer to your previous question, by Moschovakis' coding lemma we have that $\Theta$ is a high limit ordinal, and so the first high limit ordinal, under ZF+AD (unless you count $\omega$ as a high limit ordinal).
Meanwhile, nothing interesting happens when you drop choice: ZF certainly does prove that high limit ordinals exist. Consider the sequence $(\lambda_\alpha)_{\alpha\in Ord}$ given by
$\lambda_0=\omega$.
$\lambda_\eta=\sup\{\lambda_\alpha: \alpha<\eta\}$ for $\eta$ a limit.
$\lambda_{\alpha+1}=\sup\{\beta\in Ord:$ there is a surjection from $2^{\lambda_\alpha}$ to $\beta\}$ (this is just $h(\lambda_\alpha)$, in your notation).
The existence of $\lambda_\alpha$ for each $\alpha\in Ord$ follows from Replacement, once we know that $\lambda_{\alpha+1}$ exists whenever $\lambda_\alpha$ does. This latter fact might look like it requires choice, but it is in fact provable in ZF (originally I believe by Lindenbaum) using Hartog's theorem: note that if a set $A$ surjects onto an ordinal $\delta$, then $\delta$ injects into the powerset $\mathcal{P}(\alpha)$.
Note that indeed this is exactly the proof that strong limit ordinals exist: the only place we use choice in that argument is in showing that $2^\alpha$ is an ordinal when $\alpha$ is, and that's not necessary here since your definition of $h$ already ensures that the output is an ordinal.
Then for any limit $\eta>0$, $\lambda_\eta$ is by definition a high limit ordinal. And it's easy to see that $\lambda_\omega$ is the least high limit ordinal (unless, again, you count $\omega$ as a high limit ordinal).