What is the depth of recursion if we split an array into $\log_2(n)$ with each recursive call?
Let $f(n) = n/\log n$, and denote by $g(n)$ the number of applications of $f$ it takes to get $n$ below some arbitrary constant. On the one hand, $f^{(t)}(n) \geq n/\log^t n$, and so
$$
g(n) \geq \log_{\log n} n = \frac{\log n}{\log \log n}.
$$
To obtain an upper bound, notice that as long as $f^{(t)}(n) \geq \sqrt{n}$, we have $\log f^{(t)}(n) \geq (\log n)/2$, and so it takes at most $\log_{(\log n)/2} n = O(\log n/\log\log n)$ iterations to reduce $n$ to $\sqrt{n}$. The same argument applies for reducing $\sqrt{n}$ to $\sqrt[4]{n}$ and so on, and therefore
$$
g(n) \ll \frac{\log n}{\log\log n} + \frac{\log \sqrt{n}}{\log\log \sqrt{n}} + \frac{\log \sqrt[4]{n}}{\log\log \sqrt[4]{n}} + \cdots.
$$
(Here $\cdot \ll \cdot$ is the same as $\cdot = O(\cdot)$.) Notice that
$\log \sqrt{n} = \frac{1}{2} \log n$. For large $n$, we will have $\log\log \sqrt{n} \geq \frac{2}{3} \log \log n$ (say), and so
$$
\frac{\log \sqrt{n}}{\log\log \sqrt{n}} \leq \frac{2}{3} \frac{\log n}{\log\log n}.
$$
Therefore the sum is majorized by a convergent geometric series, and we conclude that
$$
g(n) = O\left(\frac{\log n}{\log\log n}\right).
$$
In total, we get that
$$
g(n) = \Theta\left(\frac{\log n}{\log \log n}\right).
$$
With more work, we can probably prove a more refined estimate such as
$$
g(n) \sim \frac{\log n}{\log \log n}.
$$