1. Given the ellipse $(C)$ with the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a > b > 0)$, an eccentricity of $\frac{\sqrt{6}}{3}$, and passing through the point $(3,-1)$, we can set up the following system of equations:

$$\begin{cases} \frac{9}{a^2} + \frac{1}{b^2} = 1\\ \frac{c^2}{a^2} = \frac{a^2 - b^2}{a^2} = \left(\frac{\sqrt{6}}{3}\right)^2 \end{cases}$$

Solving this system, we find $a^2 = 12$ and $b^2 = 4$. Thus, the equation of the ellipse $(C)$ is $\frac{x^2}{12} + \frac{y^2}{4} = 1$.

2. The equation of the line $l$ is given by $x = -2\sqrt{2}$. Let $P(-2\sqrt{2}, y_0)$ be a point on $l$, where $y_0 \in \left(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}\right)$. When $y_0 \neq 0$, let $M(x_1, y_1)$ and $N(x_2, y_2)$ be the intersection points of the line through $P$ and the ellipse $(C)$. According to the problem, $x_1 \neq x_2$.

By substituting the coordinates of points $M$ and $N$ into the equation of the ellipse, we have:

$$\begin{cases} \frac{x_1^2}{12} + \frac{y_1^2}{4} = 1\\ \frac{x_2^2}{12} + \frac{y_2^2}{4} = 1 \end{cases}$$

Subtracting these equations, we get $\frac{x_1^2 - x_2^2}{12} + \frac{y_1^2 - y_2^2}{4} = 0$. This implies that the slope of the line $MN$ is $- \frac{1}{3} \cdot \frac{x_1 + x_2}{y_1 + y_2}$. Since $PM = PN$, point $P$ is the midpoint of segment $MN$. Therefore, the slope of line $MN$ can also be written as $- \frac{1}{3} \cdot \frac{-2\sqrt{2}}{y_0} = \frac{2\sqrt{2}}{3y_0}$.

As $l' \perp MN$, the equation of line $l'$ is $y - y_0 = -\frac{3y_0}{2\sqrt{2}}(x + 2\sqrt{2})$, which simplifies to $y = -\frac{3y_0}{2\sqrt{2}}\left(x + \frac{4\sqrt{2}}{3}\right)$. Thus, line $l'$ always passes through the fixed point $\boxed{\left(-\frac{4\sqrt{2}}{3}, 0\right)}$.

If $y_0 = 0$, line $MN$ is given by $x = -2\sqrt{2}$. In this case, line $l'$ is the $x$-axis, which also passes through the point $\left(-\frac{4\sqrt{2}}{3}, 0\right)$. Therefore, $l'$ always passes through the fixed point $\boxed{\left(-\frac{4\sqrt{2}}{3}, 0\right)}$.