(1) Solution: Given that the minimum distance from any point on the ellipse to point $F(-1,0)$ is $\sqrt{2}-1$, we have the following system of equations:

$$\begin{cases} c=1 \\ a-c=\sqrt{2}-1 \end{cases}$$

Solving this system, we get:

$$\begin{cases} a=\sqrt{2} \\ c=1 \end{cases}$$

Since $b^{2}=a^{2}-c^{2}=1$, the equation of the ellipse is $\frac{x^{2}}{2}+y^{2}=1$.

(2) Proof:

$\text{(i)}$ When line $l$ is vertical, its equation is $x=-1$. We can find that $A(-1, \frac{\sqrt{2}}{2})$ and $B(-1,-\frac{\sqrt{2}}{2})$. In this case, $\overline{MA}\cdot \overline{MB}=(-1+\frac{5}{4}, \frac{\sqrt{2}}{2})\cdot (-1+\frac{5}{4},-\frac{\sqrt{2}}{2})=-\frac{7}{16}$.

$\text{(ii)}$ When line $l$ is not vertical, let the equation of line $l$ be $y=k(x+1)$. Solving the system of equations:

$$\begin{cases} y=k(x+1) \\ \frac{x^{2}}{2}+y^{2}=1 \end{cases}$$

We get the quadratic equation $(1+2k^{2})x^{2}+4k^{2}x+2k^{2}-2=0$. Let $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$, then $x_{1}+x_{2}=-\frac{4k^{2}}{1+2k^{2}}$ and $x_{1}x_{2}=\frac{2k^{2}-2}{1+2k^{2}}$.

Thus, $\overline{MA}\cdot \overline{MB}=(x_{1}+\frac{5}{4},y_{1})\cdot (x_{2}+\frac{5}{4},y_{2})=(x_{1}+\frac{5}{4})(x_{2}+\frac{5}{4})+y_{1}y_{2}$

$=x_{1}x_{2}+\frac{5}{4}(x_{1}+x_{2})+\frac{25}{16}+k(x_{1}+1)(kx_{2}+1)$

$=(1+k^{2})x_{1}x_{2}+(k^{2}+\frac{5}{4})(x_{1}+x_{2})+k^{2}+\frac{25}{16}$

$=(1+k^{2})\frac{2k^{2}-2}{1+2k^{2}}+(k^{2}+\frac{5}{4})(-\frac{4k^{2}}{1+2k^{2}})+k^{2}+\frac{25}{16}$

$=\frac{-4k^{2}-2}{1+2k^{2}}+\frac{25}{16}$

$=-2+\frac{25}{16}$

$=-\frac{7}{16}$.

Therefore, $\overline{MA}\cdot \overline{MB}$ is constant, and its value is $\boxed{-\frac{7}{16}}$.