= Solution
For $a>0$ and $n\geq2$, let $A_n(a)=\{X_n^2>a\log n\}$. Symmetry of the <standard normal distribution> and the <Mills ratio> asymptotic just proved give
$$
\mathbb P(A_n(a))=2\bigl[1-\Phi(\sqrt{a\log n})\bigr]
\sim\sqrt{\frac{2}{\pi a}}\frac{n^{-a/2}}{\sqrt{\log n}}.
$$
For $a>2$ the <series> of these <probabilities> converges, since its terms are eventually bounded by a constant times the convergent <p-series> $n^{-a/2}$. For $0<a<2$ it diverges: choose $r$ strictly between $a/2$ and one, and use $(\log n)^{-1/2}\geq n^{-(r-a/2)}$ for sufficiently large $n$, reducing to the divergent <p-series> $n^{-r}$. At $a=2$ it still diverges, as the <integral test> gives
$$
\int_2^R\frac{dt}{t\sqrt{\log t}}
=2\bigl(\sqrt{\log R}-\sqrt{\log2}\bigr)\longrightarrow\infty.
$$
For $a>2$, the first <Borel-Cantelli lemma> makes $A_n(a)$ occur only finitely often. For $0<a\leq2$, independence of the <random variables> makes the <events> $A_n(a)$ independent, so the second <Borel-Cantelli lemma> makes them occur infinitely often <almost surely>. Apply these conclusions simultaneously at $a=2+1/k$ for positive integers $k$, and at $a=2$. On the resulting <probability>-one <event>, the <limit superior> is at most every $2+1/k$ and at least two. Therefore the <Gaussian sample limsup at logarithmic scale> is
$$
\boxed{\limsup_{n\to\infty}\frac{X_n^2}{\log n}=2\quad\text{almost surely}.}
$$
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