How close is $n $ to $ n \log n$?

Well, $\log n = n^{\log_n \log n} = n^{\frac{\log \log n}{\log n}}$, so $\epsilon = \frac{\log \log n}{\log n}$.

I think the important takeaway from this is that $\epsilon$ is smaller than any constant as $n \to \infty$, since $\log\log n \ll \log n$.

For any $\epsilon > 0$, $n^{1+\epsilon}$ grows faster than $n \log n$ (in the sense that $\frac{n^{1+\epsilon}}{n \log n} \to \infty$).