Solution:

Let $O$ be the center, $s$ the side length, $a$ the apothem, and $u_i$ the unit vector from $O$ toward the midpoint of chord $A_iA_{i+1}.$ The region bounded by $\overline{PA_i},$ $\overline{PA_{i+1}},$ and the arc is the circular segment together with triangle $PA_iA_{i+1}.$ The segment has area $\frac{1}{8} - \frac{sa}{2}$ (sector minus triangle $OA_iA_{i+1}$), and the triangle at $P$ has area $\frac{s}{2}(a - P \cdot u_i),$ since $a - P \cdot u_i$ is the distance from $P$ to the chord. Adding, $$\text{area}_i = \frac{1}{8} - \frac{s}{2}\,(P \cdot u_i).$$

The given areas say $\frac{s}{2}(P \cdot u_1) = \frac{1}{8} - \frac{1}{7} = -\frac{1}{56}$ and $\frac{s}{2}(P \cdot u_3) = \frac{1}{8} - \frac{1}{9} = \frac{1}{72}.$ The normals rotate $45^\circ$ per side, so $u_3$ is $u_1$ rotated $90^\circ,$ and $u_6,$ five steps from $u_1,$ is $u_1$ rotated $225^\circ:$ $$u_6 = -\frac{\sqrt{2}}{2}\,(u_1 + u_3).$$ Therefore $$\frac{s}{2}(P \cdot u_6) = -\frac{1}{\sqrt{2}}\left(-\frac{1}{56} + \frac{1}{72}\right) = -\frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{252}\right) = \frac{\sqrt{2}}{504}.$$

The region on side $A_6A_7$ thus has area $\frac{1}{8} - \frac{\sqrt{2}}{504},$ so $n = 504.$