Your question as it stands is quite unclear, but let me take a stab at it; based on your previous questions, if nothing else I think you'll find this interesting.
The simplest interpretation of your question is to look for an analogue of the Busy Beaver function for iterates of the Turing jump through the ordinals. Such a thing does exist, although - as usual, and along similar lines to some of your previous questions - we have to index by ordinal notations instead of ordinals per se, and we run into trouble when we try to continue past the computable ordinals (see chapter $1$ of Sacks' book for the relevant background; on the other hand, see Hodes' paper on mastercodes for ideas on how to go even further):
For $l$ a notation for a computable limit ordinal, we get a sequence $(n_i)_{i\in\omega}$ of notations forming a canonical sequence for $\vert l\vert_\mathcal{O}$ (where "$\vert k\vert_\mathcal{O}$" denotes the ordinal $k$ is a notation for). The $l$th Busy Beaver function, $BB_l$, is then defined by
$$BB_l(x)=1+\sum_{i\le x}BB_{n_i}(x).$$
For $m$ a notation for a successor ordinal, we let $n$ be the canonical "predecessor notation" of $m$, and define $BB_m$ as the sum of $BB_n$ and the usual Busy Beaver function with oracle $BB_n$.
The $\mathcal{O}$-indexed "hierarchy" so defined has a number of nice properties. in particular, there is also a very tight connection with iterates of the Turing jump (as alluded to above): $BB_k$ is Turing-equivalent to $0^{(\vert k\vert_\mathcal{O})}$ for every notation $k$.
It turns out that this computation power carries over to domination: given any function $f$ dominating a Busy Beaver function with index a notation for a computable ordinal $\alpha$, we have $f\ge_T0^{(\alpha)}$.
Surprisingly, the degree $0^{(\alpha)}$ makes sense for $\alpha$ a computable ordinal, not just a notation - if $i,j$ are notations for computable ordinals, then the potentially-different sets $0^{(\vert i\vert_\mathcal{O})}$ and $0^{(\vert j\vert_\mathcal{O})}$ are of equal Turing degree.
This, in turn, lets us prove the desired domination result that $$\vert i\vert_\mathcal{O}<\vert j\vert_\mathcal{O}\implies BB_i\prec BB_j,$$ where "$\prec$" means "dominates: we argue by induction that for any notation $k$ the higher Busy Beaver function $BB_k$ dominates every function computable from $0^{(\alpha)}$ for any $\alpha<\vert k\vert_\mathcal{O}$.
This naturally suggests the question of what sets can be computed from fast-enough functions. A beautiful theorem of Solovay says that basically the situation above is the only way this can happen:
Given a set $A$, the following are equivalent: $(i)$ $A$ is hyperarithmetic, $(ii)$ there is some function $f$ such that every function dominating $f$ computes $A$.
The observation we've made about the $BB_i$s forms the $(i)$-to-$(ii)$ direction (the other direction requires a nice forcing argument). See Peter Gerdes' thesis for further work on the topic.
Emil Jeřábek answered of on mathoverflow.
Excerpt:
Essentially, yes. An old result of Kleene 1, later strengthened by Craig and Vaught 2, shows that every recursively axiomatizable theory in first-order logic without identity, and every recursively axiomatizable theory in first-order logic with identity that has only infinite models, has a finitely axiomatized conservative extension. See also Mihály Makkai’s review, Richard Zach’s summary, and a related paper by Pakhomov and Visser [3].
Let me stress that the results above apply to the literal definition of conservative extension, i.e., we extend the language of $T$ by additional predicate or function symbols, and we demand that any sentence in the original language is provable in the extension iff it is provable in $T$. If we loosen the definition so as to allow additional sorts, or extension by means of a relative interpretation, then every recursively axiomatizable first-order theory has a finitely axiomatized conservative extension.*
*There is more than one way how to show this, but for example it follows immediately from the above mentioned result that any r.e. theory with only infinite models has a finite conservative extension: simply endow the given theory with a new sort (with no further structure), and axioms that make it infinite.
Best Answer
If I understand the question correctly, it boils down to the following:
The answer to $(*)$ is, in fact, negative. Say that an ordinal $\alpha$ is fresh iff $V_\alpha\not\equiv V_\beta$ for any $\beta<\alpha$. Freshness is definable, but this means that "the supremum of the fresh ordinals" is a definable ordinal, and hence less than $\Delta$.
Consequently, we have:
$^1$Note that I'm assuming here that $\Delta$ in fact exists. While at first glance a pretty banal statement for "Platonic $V$," there is some surprising nuance here; that said, in the context of this question I think it's worth largely taking for granted.