Let $P$ be the convex polygon in the complex plane whose vertices are the roots of
\[z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0.\]The area of $P$ can be expressed in the form $\frac{a + b \sqrt{c}}{d},$ where $a,$ $b,$ $c,$ $d$ are positive integers, in simplest form.  Find $a + b + c + d.$
Multiplying the given equation by $z - 1,$ we get
\[(z - 1)(z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1) = 0,\]or $z^8 = 1.$  Thus, the vertices of $P$ are the eighth roots of unity, other than 1.

[asy]
unitsize (2 cm);

pair A, B, C, D, E, F, G, O;

A = dir(45);
B = dir(90);
C = dir(135);
D = dir(180);
E = dir(225);
F = dir(270);
G = dir(315);
O = (0,0);

filldraw(A--B--C--D--E--F--G--cycle,gray(0.7));
//draw((-1.2,0)--(1.2,0));
//draw((0,-1.2)--(0,1.2));
draw(Circle((0,0),1),red);
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
draw(O--E);
draw(O--F);
draw(O--G);
[/asy]

We can dissect the polygon into six isosceles triangles, where the equal sides have length 1 and the angle between them is $45^\circ,$ and one isosceles triangle, where the equal sides have length 1, and the angle between them is $90^\circ.$  Thus, the area of polygon $P$ is
\[6 \cdot \frac{1}{2} \cdot 1^2 \cdot \sin 45^\circ + \frac{1}{2} = \frac{1 + 3 \sqrt{2}}{2}.\]The final answer is $1 + 3 + 2 + 2 = \boxed{8}.$