I'm going to give it a shot -- Borel-Cantelli is a tad hazy in my mind, though, so you definitely should doublecheck.
Fix any $\epsilon > 0$, and define the event $$E_n\stackrel{\rm{}def}{=} \mathbb{1}_{\left\{\frac{\lvert X_n\rvert}{n}\geq \epsilon\right\}}$$
so that
$$ \mathbb{P} E_n = \mathbb{P}\left\{ \lvert X_n\rvert \geq n\epsilon \right\} = \operatorname{erfc}\left(\frac{\epsilon n}{ \sqrt{2} }\right) \operatorname*{\sim}_{n\to\infty} \sqrt{\frac{2}{\pi}} \frac{e^{-\frac{\epsilon^2n^2}{2}}}{\epsilon n}$$
and thus $\sum_{n=1}^\infty \mathbb{P} E_n < \infty$. By Borel-Cantelli,
$$ \mathbb{P}\left(\limsup_{n\to\infty} E_n\right) = 0 $$
that is,
$$ \forall \epsilon > 0, \forall_{\rm{}a.s.} \omega\in\Omega,\ \exists N\geq 0,\forall n\geq N,\ \frac{\lvert X_n\rvert}{n}< \epsilon$$
or, equivalently, $\displaystyle\lim_{n\to\infty}\frac{\lvert X_n\rvert}{n}\to0$ a.s.
This also seems to work for $\frac{X_n}{\ln n}$, as $$\sum_{n=1}^\infty\frac{e^{-\frac{\epsilon^2}{2}\ln^2n}}{\epsilon \ln n} < \infty.$$