Does $\sum_{n=4}^\infty \left(\frac{1}{\log(\log(n))}\right)^{\log(n)}$ converge?

Your argument works more generally.

Given any positive function $f(n):$

$$f(n)^{\log n}=e^{\log f(n)\log n}=n^{\log f(n)}$$

Now, if $f(n)\to\infty$ as $n\to\infty$ we can get that $$\sum \left(\frac1{f(n)}\right)^{\log n}$$ converges, by your reasoning.

Even weaker still, if there is some $c>e$ and some $N,$ such that $f(n)\geq c$ for all $n\geq N,$ then the series converges.

So you only need $f(n)=\log\log n\geq 3$ for $n$ sufficiently large, say $n>e^{e^3}.$

If $f$ is non-decreasing, the series diverges if and only if $f(n)\leq e$ for all $n.$

By Cauchy condensation test we have

$$\sum_{n=4}^\infty 2^n\left(\frac{1}{\log(\log(2^n))}\right)^{\log(2^n)}=\sum_{n=4}^\infty \left(\frac{e}{\log(n\log 2)}\right)^{n\log 2}$$

which converges.

Following the approach outlined in the OP, we have

$$\begin{align} \left(\frac{1}{\log(\log(n))}\right)^{\log(n)}&=e^{-\log(n)\log(\log(\log(n)))}\\\\ &=n^{-\log(\log(\log(n)))} \end{align}$$

For all $n>e^{e^e}$, $\log(\log(\log(n)))>1$ and inasmuch as $\log(n)$ montotonically increases, the series converges.