Pick parameter function that minimises whole function
This is probably not possible. Let $n_1,\ldots,n_k$ be the sequence of $n$s encountered in the recurrence, starting with $n_1 = n$. Using the concavity of $\log$ (and eliminating the big $O$), we get that
$$ T(n) \geq k \log \frac{f(n_1) + \cdots + f(n_k)}{k}. $$
Assuming $f$ is monotone, we have $k \geq \log n_1/\log (n_1/n_2) = \log n/\log f(n)$. This monotonicity also shows that $n_{(k+1)/2} \geq \sqrt{n}$. Therefore
$$ T(n) \geq \log n \frac{\log f(\sqrt{n}) - 1}{\log f(n)}. $$
For all $f(n)$ I can think of, $\log f(\sqrt{n})/\log f(n) = \Theta(1)$, so they can't give you anything better than $\log n$. This also says that if you're looking for a monotone $f$ which beats $\log n$, then it must satisfy $\log f(\sqrt{n}) = o(\log f(n))$.