How can I use the Central Limit Theorem to calculate the distribution of $\bar{X}$? The central limit theorem says that $$
\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}} \stackrel{\mathcal{D}}{\rightarrow} N(0,1)
$$
What is the distribution of $\bar{X}$? I've seen it given as $\sum X \sim N\left((n)\left(\mu_x\right),(\sqrt{n})\left(\sigma_x\right)\right)$, but don't know how to prove this.

There's probably an easier way to prove this, simply based on linearity of expectation and variance, and the fact that Gaussians are defined by these two moments, however:
CLT implies convergence by distribution (also called weak convergences), which means that the Cumulative Density Function converges for $n \rightarrow \inf$
This means that for $Y = \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}},$ the CDF will converge to the one of $N(0,1)$, which is:
$$F_Y \rightarrow \frac{1}{2}\left[1+\text{erf}\left(\frac{y}{\sqrt{2}}\right)\right]$$
Now if we call $Z = \sum X = n\bar{X}$, we have that:
\begin{align}F_Z &\rightarrow \frac{1}{2}\left[1+\text{erf}\left(\frac{\frac{Z}{n}-\mu}{\sqrt{2}\sigma}\sqrt{n}\right)\right] \\&= \frac{1}{2}\left[1+\text{erf}\left(\frac{Z-n\mu}{\sqrt{2}\sigma\sqrt{n}}\right)\right] \end{align}
which is the CDF of $N(n\mu, \sigma\sqrt{n}).$
