Given an initial value of $n = 2^i$, the argument is gradually reduced so that after one stage it is $2^{i - \log i}$, and so forth. If we examine the exponent alone, it satisfies
$$
i(0) = \log_ n,  i(k+1) = i(k) - \log i(k)
$$

We approximate this difference equation with the differential equation $$
y(0) = \log_2 n,
y'(x) = -\log_2 y(x)
$$
yielding
$$
y(x) = \log_2 n - (\ln 2) \text{li}(x)
$$
where $li$ is the logarithmic integral.

The value $T(n)$ is the value of $x$ such that $y(x) = 1$, that is, such that $\text{li} (x) = (\log_2 n - 1)/\ln 2$.

For $n$ sufficiently large we have $li(x) \sim x/\ln x$. Hence we should have
$$
x \sim (\ln 2)^{-2} \ln n \ln \ln n
$$
that is,
$$
T(n) \sim (\log_2 n) (\log_2 \log_2 n)
$$

Obviously I have left out a lot of details but this be basically right...