Let $P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)$. What is the minimum perimeter among all the $8$-sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)$?
First, we can factor $P(z) = z^8 + (4 \sqrt{3} + 6) z^4 - (4 \sqrt{3} + 7)$ as
\[P(z) = (z^4 - 1)(z^4 + 4 \sqrt{3} + 7).\]The solutions to $z^4 - 1 = 0$ are 1, $-1,$ $i,$ and $-i$.

If $z^4 + 4 \sqrt{3} + 7 = 0,$ then
\[z^4 = -4 \sqrt{3} - 7 = (-1)(4 \sqrt{3} + 7),\]so $z^2 = \pm i \sqrt{4 \sqrt{3} + 7}.$

We try to simplify $\sqrt{4 \sqrt{3} + 7}.$  Let $\sqrt{4 \sqrt{3} + 7} = a + b.$  Squaring both sides, we get
\[4 \sqrt{3} + 7 = a^2 + 2ab + b^2.\]Set $a^2 + b^2 = 7$ and $2ab = 4 \sqrt{3}.$  Then $ab = 2 \sqrt{3},$ so $a^2 b^2 = 12.$  We can then take $a^2 = 4$ and $b^2 = 3,$ so $a = 2$ and $b = \sqrt{3}.$  Thus,
\[\sqrt{4 \sqrt{3} + 7} = 2 + \sqrt{3},\]and
\[z^2 = \pm i (2 + \sqrt{3}).\]We now try to find the square roots of $2 + \sqrt{3},$ $i,$ and $-i.$

Let $\sqrt{2 + \sqrt{3}} = a + b.$  Squaring both sides, we get
\[2 + \sqrt{3} = a^2 + 2ab + b^2.\]Set $a^2 + b^2 = 2$ and $2ab = \sqrt{3}.$  Then $a^2 b^2 = \frac{3}{4},$ so by Vieta's formulas, $a^2$ and $b^2$ are the roots of
\[t^2 - 2t + \frac{3}{4} = 0.\]This factors as $\left( t - \frac{1}{2} \right) \left( t - \frac{3}{2} \right) = 0,$ so $a^2$ and $b^2$ are equal to $\frac{1}{2}$ and $\frac{3}{2}$ in some order, so we can take $a = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $b = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}.$  Hence,
\[\sqrt{2 + \sqrt{3}} = \frac{\sqrt{2} + \sqrt{6}}{2} = \frac{\sqrt{2}}{2} (1 + \sqrt{3}).\]Let $(x + yi)^2 = i,$ where $x$ and $y$ are real numbers.  Expanding, we get $x^2 + 2xyi - y^2 = i.$  Setting the real and imaginary parts equal, we get $x^2 = y^2$ and $2xy = 1.$  Then $4x^2 y^2 = 1,$ so $4x^4 = 1.$  Thus, $x = \pm \frac{1}{\sqrt{2}},$ and the square roots of $i$ are
\[\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i = \frac{\sqrt{2}}{2} (1 + i), \ -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i = -\frac{\sqrt{2}}{2} (1 + i).\]Similarly, we can find that the square roots of $-i$ are
\[\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i = \frac{\sqrt{2}}{2} (1 - i), \ -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i = \frac{\sqrt{2}}{2} (-1 + i).\]Hence, the solutions to $z^4 = -4 \sqrt{3} - 7$ are
\[\frac{1 + \sqrt{3}}{2} (1 + i), \ -\frac{1 + \sqrt{3}}{2} (1 + i), \ \frac{1 + \sqrt{3}}{2} (1 - i), \ \frac{1 + \sqrt{3}}{2} (-1 + i).\]We plot these, along with 1, $-1,$ $i,$ $-i$ in the complex plane.

[asy]
unitsize(2 cm);

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

A = (1,0);
B = (-1,0);
C = (0,1);
D = (0,-1);
E = (1 + sqrt(3))/2*(1,1);
F = (1 + sqrt(3))/2*(-1,-1);
G = (1 + sqrt(3))/2*(1,-1);
H = (1 + sqrt(3))/2*(-1,1);

draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));
draw(A--C--B--D--cycle,dashed);
draw(A--E--C--H--B--F--D--G--cycle,dashed);

dot("$1$", A, NE, fontsize(10));
dot("$-1$", B, NW, fontsize(10));
dot("$i$", C, NE, fontsize(10));
dot("$-i$", D, SE, fontsize(10));
dot("$\frac{1 + \sqrt{3}}{2} (1 + i)$", E, NE, fontsize(10));
dot("$-\frac{1 + \sqrt{3}}{2} (1 + i)$", F, SW, fontsize(10));
dot("$\frac{1 + \sqrt{3}}{2} (1 - i)$", G, SE, fontsize(10));
dot("$\frac{1 + \sqrt{3}}{2} (-1 + i)$", H, NW, fontsize(10));
[/asy]

The four complex numbers 1, $-1,$ $i,$ $-i$ form a square with side length $\sqrt{2}.$  The distance between $\frac{1 + \sqrt{3}}{2} (1 + i)$ and 1 is
\begin{align*}
\left| \frac{1 + \sqrt{3}}{2} (1 + i) - 1 \right| &= \left| \frac{-1 + \sqrt{3}}{2} + \frac{1 + \sqrt{3}}{2} i \right| \\
&= \sqrt{\left( \frac{-1 + \sqrt{3}}{2} \right)^2 + \left( \frac{1 + \sqrt{3}}{2} \right)^2} \\
&= \sqrt{\frac{1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3} + 3}{4}} \\
&= \sqrt{2}.
\end{align*}Thus, each "outer" root has a distance of $\sqrt{2}$ to its nearest neighbors.  So to the form the polygon with the minimum perimeter, we join each outer root to its nearest neighbors, to form an octagon with perimeter $\boxed{8 \sqrt{2}}.$