The answer is no.
Here is the counter-example.
First, construct $(X,Y)$ such that $X$ and $Y$ follows the normal distribution with zero covariance but $(X,Y)$ is not jointed normal distribution. Suppose
$$X \sim \mathcal{N}(0,\sigma_1^2)$$
$$Y = \frac{\sigma_2}{\sigma_1}\cdot X\cdot B$$
where $B$ is independent to $X$ and with the law $\mathbb{P}(B = -1) =\mathbb{P}(B = 1) = \frac{1}{2}$.
Then, $Y$ follows the normal distribution $ \mathcal{N}(0,\sigma_2^2)$, indeed it suffices to compute the characteristic function of $Y$:
$$\begin{align}
\mathbb{E}\left(e^{itY}\right)&= \mathbb{E}\left(e^{it\frac{\sigma_2}{\sigma_1}\cdot X\cdot B}\right)\\
&= \mathbb{E}\left(\mathbb{E}\left(e^{it\frac{\sigma_2}{\sigma_1}\cdot X\cdot B}|B\right)\right)\\
&= \mathbb{E}\left(e^{-\frac{1}{2}\sigma_1^2\left(t\cdot \frac{\sigma_2}{\sigma_1}B\right)^2}\right)\\
&= \mathbb{E}\left(e^{-\frac{1}{2}\sigma_2^2t^2B^2}\right)\\
&= \mathbb{E}\left(e^{-\frac{1}{2}\sigma_2^2t^2}\right)\\
\end{align}$$
which is exactly the characteristic function of $ \mathcal{N}(0,\sigma_2^2)$. Then, $Y \sim \mathcal{N}(0,\sigma_2^2)$.
It's easy to prove that
$$Cov(X,Y) = \mathbb{E}(XY) = \frac{\sigma_2}{\sigma_1}\mathbb{E}(X^2B) = \frac{\sigma_2}{\sigma_1}\mathbb{E}(X^2)\underbrace{\mathbb{E}(B)}_{=0} = 0$$
However, as $X+ \frac{\sigma_1}{\sigma_2} Y = X(B+1)$ does not follow the normal distribution (as $\mathbb{P}(X(B+1)=0) =\frac{1}{2}$), $(X,Y)$ is not then joint normal distribution.
Now, return back to your question, it suffices to denote
$$X_n = X$$
$$Y_n = Y$$
and you have the counter-example.
