The definition of factorial gives $ \log(n!)-\log((n-1)!)=\log(n)\tag{1} $ Since the derivative of $\log(x)$ is $1/x$, $(1)$ and the Mean Value Theorem yield $ \begin{align} \log(\log(n!))-\log(\log((n-1)!)) &\in\left(\frac{\log(n)}{\log(n!)},\frac{\log(n)}{\log(n!)-\log(n)}\right)\\ &=\frac{\log(n)}{n\log(n)-n+O(\log(n))}\tag{2} \end{align} $ The density of $\log(\log(n!))$ is the reciprocal of $(2)$: $n-\frac{n}{\log(n)}+O(1)$ and $ \log(\log(n!))=\log(n)+\log(\log(n))-\frac{1}{\log(n)}+O\left(\frac{1}{\log(n)^2}\right)\tag{3} $ As $n\to\infty$, $\log(\log(n!))\sim\log(n)$ and the density $\sim n$. Since the limiting density is determined when $n$ is large, we get that the density of $\{\log(\log(n!))\}$ is $\frac{e^x}{e-1}$ on $[0,1]$.
More explanation about $\frac{e^x}{e-1}$:
Since $\log(\log(n!))=\log(n)+\log(\log(n))-\frac{1}{\log(n)}+O\left(\frac{1}{\log(n)^2}\right)$, $\{\log(\log(n!))\}$ cycles through $[0,1]$ just a bit quicker than $\log(n)$ does; approximately when $n$ goes to $ne\left(1-\frac{1}{\log(n)}\right)$. In each of those cycles, the logarithm of the density, $\log\left(n-\frac{n}{\log(n)}\right)+O\left(\frac{1}{n}\right)$, increases approximately linearly by $1$. Thus, the density of $\{\log(\log(n!))\}$ is proportional to $e^x$, and $\frac{e^x}{e-1}$ is normalized to have total weight $1$.
Density Details:
Let $I_n=\{k\in\mathbb{Z}:n-1<\log(\log(k!))\le n\}$. The density approximated here is the function $\phi:[0,1]\mapsto\mathbb{R}$ so that $ \int_a^b\phi(x)\;\mathrm{d}x=\lim_{n\to\infty}\left.\left|\{k\in\mathbb{Z}:n-1+a<\log(\log(k!))\le n-1+b\}\right|\middle/\left|I_n\right|\right. $ Within a given $I_n$ this density is roughly proportional to the reciprocal of the distance between $\log(\log((k-1)!))$ and $\log(\log(k!))$, which is $k-\frac{k}{\log(k)}+O(1)$. For $k\in I_n$, let $x=\log(\log(k!))-n+1=\log(k)+\log(\log(k))-n+1+O\left(\frac{1}{\log(k)}\right)$. Then $ k=\frac{e^{x+n-1+O(1/(x+n))}}{(x+n-1)^{1-1/(x+n)}}=\frac{e^{n-1+O(1/(x+n))}}{(x+n-1)^{1-1/(x+n)}}e^x $ Thus, in terms of $x$, the density is $ k-\frac{k}{\log(k)}+O(1)=\frac{e^{n-1+O(1/(x+n))}}{(x+n-1)^{1-1/(x+n)}}e^x $ As $n\to\infty$, the coefficient of $e^x$ tends toward constancy. Normalizing this so that the integral over $[0,1]$ is $1$, we get $\phi(x)=\frac{e^x}{e-1}$.
We get the same density if we consider $\displaystyle\lim_{N\to\infty} \bigcup_{n\le N} I_n$. However, if we use a partial $I_N$, the fact that $|I_N|$ is approximately $(e-1)\left|\bigcup_{n causes bad behavior.
Thus, the fractional parts of $\log(\log(n!))$ are dense in $[0,1]$, but not uniformly distributed.
Charts
On $I_{10}$, the counts in each interval of size $0.01$ are

On $I_{11}$, the counts in each interval of size $0.01$ are

On $I_{12}$, the counts in each interval of size $0.01$ are

On $I_{13}$, the counts in each interval of size $0.01$ are

So as described above, on each $I_n$, the density is $\frac{e^x}{e-1}$. However, since the counts on $I_{n+1}$ are approximately $e$ times the counts on $I_n$, the picture is different if we don't consider complete intervals $I_n$:

The jump has a ratio of about $e\left(1-\frac1{\log(n)}\right)$ since $\{\log(\log(n!))\}$ cycles through $[0,1]$ as $n$ goes to $ne\left(1-\frac1{\log(n)}\right)$, as mentioned above.