Compressing information about the halting problem for oracle Turing machines
Let $J^A(e)$ be the output of the $e$th Turing machine equipped with oracle $A$, on input $e$. Here $J$ stands for "jump". (In case of non-halting, $J^A(e)$ is undefined.)

An oracle $A$ is jump-traceable if there is a computable nondecreasing function $h:\mathbb N\to\mathbb N$ such that for all $e$, $J^A(e)\in T_e$ for some computably enumerable family of finite sets $(T_e)_{e\in\mathbb N}$ with $|T_e|\le h(e)$ for all $e$.

Let's consider $f^A(k,n)=$the string of length $n$ indicating which are the first $k$ of the Turing machines numbers $0,\dots,n-1$ to halt.
(If fewer than $k$ halt, $f^A(k,n)$ is undefined.)

Note that $f^A$ is partial computable relative to $A$. Thus there is a computable function $g$ such that $f^A(k,n)=J^A(g(k,n))$.

To compress the first $n$ bits of the halting set for $A$ it suffices to say which one of of the $h(e)$ many elements of $T_e$ is the right one, where $e=g(k,n)$ and $k$ is the correct number of halting TMs.

There is a proper hierarchy of jump-traceable oracles (Nies, Computability and Randomness, Theorem 8.5.2). So by choosing $h$ appropriately small, we get a candidate for an oracle $A$ as you asked for.

It is a reasonably good candidate in the sense that we have one direction (the upper bound on the growth rate) and that provably the method by which we obtained the upper bound does not give any much smaller upper bound than that.