Note that $n^{\log(n)}=e^{\log(n)^2}$ and $\log(n)^n=e^{n\log(\log(n))}$. Thus $a_n=e^{\log(n)^2-n\log(\log(n))},$ and $a_n\rightarrow0$ iff $\log(n)^2-n\log(\log(n))\rightarrow-\infty$ (which I'm pretty sure it does). If you can find a sequence which bounds $\log(n)^2-n\log(\log(n))$ from above and also goes to $-\infty$ fast enough, you should be able to prove that the sum converges.
(this is for the previous version with $\log(n^n)$ instead of $\log(n)^n$)
If $a_n\not\rightarrow0$, then the series $\sum a_n$ must diverge. Note that $a_n=f(n)$ where $f(x)=\frac{x^{\log(x)}}{\log(x^x)}=\frac{x^{\log(x)}}{x\log(x)}=\frac{x^{\log(x)-1}}{\log(x)}.$ Thus, if we show that $\lim_{x\rightarrow\infty}f(x)\neq0$, then the series must diverge.
We have that $\lim_{x\rightarrow\infty}\frac{x^{\log(x)-1}}{\log(x)}=\frac{\infty}{\infty}$ so using L'Hopital this equals $\lim_{x\rightarrow\infty}\frac{(\log(x)-1)x^{\log(x)-2}\cdot\frac{1}{x}}{\frac{1}{x}}=\lim_{x\rightarrow\infty}(\log(x)-1)x^{\log(x)-2}=\infty$