As explained in [my answer to your previous question](https://mathoverflow.net/a/283778/8133), 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](https://en.wikipedia.org/wiki/Hartogs_number): 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).