Let $Z_n := Z := g(X) := n X^n$. The relation between $Z$ and $Y$ is quite simple (one is a scaled version of the other). Let's figure out the distribution of $Z$. Assume for simplicity that $n$ is odd. Then, $g^{-1}(z) = (z/n)^{1/n}$. Hence, $|{g^{-1}}'(z)| = \frac1n (|z|/n)^{1/n-1}$.
It follows that the density of $Z$ is
$$
f_{Z_n}(z) = f_X(g^{-1}(z)) |{g^{-1}}'(z)| = \frac{n^{1/n}}{\sqrt{2\pi}} |z|^{\frac1n-1} e^{-(z^{1/n} n^{-1/n} - \mu)^2/(2\sigma^2)}
$$
Since $n^{1/n} \to 1$ and $|z|^{1/n} \to 1$ for all $z \neq 0$, the density pointwise converges to
$$
f_{Z_n}(z) \to \frac{1}{2\pi} e^{-(1-\mu)^2/2(\sigma)^2} |z|^{-1}, \quad z \neq 0
$$
which might be a good approximation for large $n$ (EDIT: it turns out it is not in general! See the edit.). Note, however, that this limit is not a density since it does not integrate to something finite. (Formally, one might be able to show that the distribution of $Z_n$ converges to a point mass at zero, in "some" sense. However, there is some mass that escapes to infinity, anything above 1 basically. So needs some care and the result might not be true. Informally, the limiting distribution is a mixture of a point mass at 0 and two point masses at $\pm \infty$ for $n$ odd.)
EDIT: Assume $n$ is odd. To clarify the situation, Let $X_1 := X\cdot 1_{\{|X| \le 1\}}$ and $X_2 := X\cdot 1_{\{|X| > 1\}}$ where $1_{\{|X| \le 1\}}$ is 1 if $|X| \le 1$ and zero otherwise and similarly for the other indicator function. We have $X = X_1 + X_2$ with $|X_1| \le 1$ almost surely and $|X_2| > 1$ almost surely. It is not hard to see that
$$
Y_n := X^n = (X_1 + X_2)^n = X_1^n + X_2^n.
$$
with $|X_1^n| \le 1$ and $|X_2^n| > 1$ a.s.
Assuming that $X$ has a continuous distribution (so that $\mathbb P(X=1) = 0$), it is not hard to see that $X_1^n$ converges in distribution to a point mass at 0. However, $X_2^n$ does not converge in distribution. In fact, it should be straightforward to show that
$$ P(X_2^n \in (0,t)) \to 0, \quad \text{as}\; n\to \infty$$ for any finite $t > 0$. This is what can be informally described as "the mass in the distribution of $X_2^n$ is escaping to infinity".
Since we also have $\mathbb P( X_1^n \in (s,\infty)) \to 0$ for any $s > 0$, it follows that $\mathbb P(Y_n \in (s,t)) \to 0$ for any $t > s > 0$. The only intervals that will have positive mass in the limit are those that contain 0. That is, if $I$ is an interval with $0 \in I$, then
$$
P(Y_n \in I) \to P(|X| \le 1), \quad \text{as}\; n \to \infty.
$$
Otherwise (that is, if $0 \notin I$), we have $\mathbb P(Y_n \in I) \to 0$.
