Consider the following definition.
A sequence of random variables ${(X_n)}_{n \in \mathbb{N}}$ is called asymptotically normally distributed if there exist sequences $\mu_n \in \mathbb{R}$ and $\sigma_n: 0 < \sigma_n < \infty$ such that
$$ \frac{1}{\sigma_n}(X_n - \mu_n) \overset{d}{\rightarrow} \mathscr{N}(0,1). $$ In this case one writes $X_n \sim AN(0,1)$.
I was wondering whether the above implies that there exists $n_0 \in \mathbb{N}$ such that all the random variables $X_n$ for $n > n_0$ will be normally distributed with parameters $(\mu_n, \sigma_n^2)$?
Reproducing and expanding my comment above:
Well, no, the same way that for sequences having $\lim_{n\to\infty} a_n=a$ does not imply the existence of some $N$ such that $a_n=a$ for every $n\geq N$. In your case, consider the "usual" Central Limit Theorem for easy counterexamples.
For instance, consider $(Y_n)_n$ to be a sequence of i.i.d. Rademacher random variables (uniform on $\{-1,1\}$). Then $X_n \stackrel{\rm def}{=} \sum_{k=1}^n Y_k$ defines a sequence of random variables, and by the CLT we have $$ \frac{X_n}{\sqrt{n}}\xrightarrow[n\to\infty]{d} \mathcal{N}(0,1) $$ so (taking $\sigma_n=\sqrt{n}$ and $\mu_n=0$ for all $n$) $(X_n)_n$ is asymptotically normally distributed. However, for every $n\geq 1$ we have that $X_n$ is discrete, so definitely not normally distributed (actually at statistical distance $1$ from Gaussian).