Let's Do Geometry.

Draw lines parallel to one side of the square that passes through the vertices of the octagons as shown. Label four adjacent lines passing through adjacent vertices of the octagons $A, B, C \text{and} D$ .

The distance from line $B$ to line $C$ is the length of the side of one of the octagons, so it is $2$ . the distance from $C$ to $D$ is the length of one leg of an isosceles right triangle with hypotenuse of length $2$ . This length is $\frac{2}{\sqrt{2}} = \sqrt{2}$ .

It follows that the side of the square has a length is $(4\times 2)+ (5\times\sqrt{2}) =8 +5\sqrt{2}.$

$\implies\text{ its area is} (8+5\sqrt{2})^2= 64 + 80\sqrt{2} +50 = 114+80\sqrt{2} \implies m= 114 \text{ and } n = 80 \implies m+n= \boxed{194}.$

The side of the square = ((4+2√2)/√2) + 2 + (6+4√2)/√2 = 8+5√2. The area of the square = (8+5√2)^2 = 114 + 80√2. m+n = 194