Let $X_n$ be a sequence of random variables converging weakly to $X$, which is a random variable having a Gaussian distribution $\mathcal{N}(0,1)$.
Let $a_n$ be an increasing sequence. What do I need in order to be true that $$\lim\limits_{n \rightarrow \infty}\, P(X_n > a_{n} ) \, / \, P(X> a_n) \, = \, C \, \geq \, 0 \, $$ i.e., $C \neq \infty$. Are $P(X_n > a_{n} ) $ and $P(X> a_n)$ always of the same order?
If $X$ is a standard Gaussian random variable, then we have $$\mathbb P\{X>\sqrt 2t\} \overset{t\to +\infty}{\sim}\frac Ct\exp\left(-t^2\right),$$ where $C$ is independent of $t$.
In the context of the question, one can define $X_n:=X-\sqrt 2c/n$ and $a_n:=\sqrt 2n$, where $c\in\mathbb R$. We have $$P(X_n > a_{n} ) \, / \, P(X> a_n)\sim\frac{\exp\left(-(n+c/n)^2\right)}{n+c/n} \frac{n}{\exp\left(-n^2\right)} \to \exp(-2c) $$ Choosing instead $X_n:=X+\sqrt 2\log n/n$ and the same $a_n$ shows that we may have $P(X_n > a_{n} ) \, / \, P(X> a_n)\to +\infty$.
Choosing $X_n:=X+\sqrt 2/\sqrt nn$ and the same $a_n$ shows that we may have $P(X_n > a_{n} ) \, / \, P(X> a_n)\to 0$.