@GEdgar in a comment suggested that this is not known, and @Peter suggested that someone work out how it follows from

Schanuel's Conjecture: If $z_1,\dots,z_k$ are linearly independent over $\mathbb Q$, then $\mathbb Q(z_1,\dots,z_k,e^{z_1},\dots,e^{z_k})$ has transcendence degree at least $k$ over $\mathbb Q$.

So here we go:

**Proposition**: Schanuel's Conjecture implies that $\ln(\ln(n))\not\in\mathbb Q$ for all $n\in\mathbb N$ with $n>1$.

**Proof**:
Suppose $\ln(\ln(n))=r\in\mathbb Q$, so that $e^{e^r}=n$.
Let $z_1=r$ and $z_2=e^r$.
Suppose
$$az_1+bz_2=0,\qquad a,b\in\mathbb Q.$$
Then either $b=0$ or $e^r=-az_1/b\in\mathbb Q$. But by Schanuel's Conjecture with $k=1$, we have that $e^r$ is transcendental. So it must be that $b=0$. If $z_1=0$ then $e^{e^0}=n$ which is false (simply since $2<e<3$). We conclude that $(a,b)=(0,0)$ and so we have shown that $z_1, z_2$ are linearly independent over $\mathbb Q$.

So by Schanuel's Conjecture,
$$\mathbb Q(z_1,z_2,e^{z_1},e^{z_2})=\mathbb Q(r,e^r,e^r,e^{e^r})=\mathbb Q(e^r,e^{e^r})$$
has transcendence degree at least 2 over $\mathbb Q$. But this implies that both $e^r$ and $e^{e^r}$ are transcendental numbers. $\Box$