What is the minimum complexity oracle that separates PSPACE from the polynomial hierarchy?
I believe if you trace through the argument given, e.g., in Section 4.1 of Ker-I Ko's survey, you get an upper bound of $\mathsf{DTIME}(2^{2^{O(n^2)}})$. In fact, we can replace $n^2$ here with any function $nf(n)$ where $f(n) \to \infty$ as $n \to \infty$. This isn't quite what was asked for, but it's close.

In particular, using the translation between oracle separations and circuit lower bounds, and following Ko's notation, we have the following:


We will diagonalize over strings of length $t(n) = p_n(m(n))$ where $p_n(x) = x^n + n$ is "the" $n$-th polynomial (in some enumeration of poly-time algorithms) and $m(n)$ will be specified below.
Translating into circuit lower bounds, this means we're considering bounded-depth circuits on $2^{t(n)}$ inputs.
The requirement (see p. 15 of Ko) we need $m(n)$ to satisfy $\frac{1}{10} 2^{m/(d-1)} &gt; d p_n(m(n))$ for all $n$. Here $d$ is the depth of the circuits we want to diagonalize against, or equivalently the level $\mathsf{\Sigma_d^p}$ of $\mathsf{PH}$ we want to diagonalize against. To diagonalize against all of $\mathsf{PH}$, simply choose $d$ to be a function of $n$ that is $\omega(1)$; we may choose such a $d$ that grows arbitrarily slowly, though (perhaps subject to some computability assumption on $d(n)$, but that should be no obstacle). If we make the guess that $d(n)$ is constant (even though it's not, but it will grow arbitrarily slowly), then we see that $m(n)$ around $2^n$ should work.
This means that $t(n) \sim 2^{n^2}$, so we are looking for a lower bound against circuits with $\sim 2^{2^{n^2}}$ inputs.
Trevisan and Xue (CCC '13) showed that one can find an assignment on which a given bounded-depth circuit on $N$ inputs doesn't compute PARITY with a seed of $polylog(N)$ length.
For us $N=2^{2^{n^2}}$, so $polylog(N) = 2^{O(n^2)}$. We can brute force over such seeds in $2^{2^{O(n^2)}}$ time and use the first one that works.


To replace the $n^2$ with $nf(n)$, just let $p_n(x) = x^{f(n)} + f(n)$ instead.

Interestingly, if I'm understanding correctly, I believe this implies that if one could improve the Trevisan-Xue...


...to a pseudodeterministic/Bellagio algorithm (see Andrew Morgan's comment below), one would get that $\mathsf{BPEXP} \not\subseteq \mathsf{P/poly}$; or
...to a nondeterministic algorithm that guessed $polylog(N)$ bits but then ran in $poly(N)$ time, and such that on any accepting path it makes the same output (cf. $\mathsf{NPSV}$), it would imply $\mathsf{NEXP} \not\subseteq \mathsf{P/poly}$; or
... to a deterministic algorithm, one would get $\mathsf{EXP} \not\subseteq \mathsf{P/poly}$.


On the one hand, this suggests why derandomizing the switching lemma further should be hard - an argument which I'm not sure was known before! On the other hand, this strikes me as a kind of interesting take on hardness versus randomness (or is this actually a new thing, oracles versus randomness?).