1. Since the left and right focal points are $F\_1$ and $F\_2$ respectively, and the endpoints of the minor axis are $A$ and $B$, and the quadrilateral $F\_1AF\_2B$ is a square with an edge length of $2$, we have $a = 2$ and $b = c$ (where $c$ is the distance from the center to a focus). Also, $a^2 = b^2 + c^2$ which gives $b^2 = 2$. Hence, the equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$.

2. Let $C=(-2,0)$ and $D=(2,0)$. Assume $M=(2,y\_0)$ and $P=(x\_1, y\_1)$. Then, $\overrightarrow{OP} = (x\_1, y\_1)$ and $\overrightarrow{OM} = (2, y\_0)$.

The equation of the line $CM$ is $y - 0 = \frac{y\_0}{4}(x + 2)$, which simplifies to $y = \frac{y\_0}{4}x + \frac{1}{2}y\_0$.

Substituting this into the equation of the ellipse $x^2 + 2y^2 = 4$, we obtain $(1 + \frac{y\_0^2}{8})x^2 + \frac{1}{2}y\_0^2x + \frac{1}{2}y\_0^2 - 4 = 0$.

The two roots of this equation are $-2$ and $x\_1$. By Vieta's theorem, we have $x\_1 - 2 = -\frac{4y\_0^2}{y\_0^2+8}$, which simplifies to $x\_1 = \frac{-2y\_0^2 + 16}{y\_0^2 + 8}$. Consequently, $y\_1 = \frac{8y\_0}{y\_0^2 + 8}$.

Thus, $\overrightarrow{OP} = (\frac{-2y\_0^2 + 16}{y\_0^2 + 8}, \frac{8y\_0}{y\_0^2 + 8})$.

Now, $\overrightarrow{OM} \cdot \overrightarrow{OP} = \frac{8y\_0^2 + 32}{y\_0^2 + 8} + \frac{8y\_0^2}{y\_0^2 + 8} = \frac{4y\_0^2 + 32}{y\_0^2} = 4$.

Therefore, $\boxed{\overrightarrow{OM} \cdot \overrightarrow{OP} = 4}$ is a constant value.