We first note that we can consider only positive values of $a, b$ and $c$, since the absolute values of the variables are calculated in all instances (both as the absolute values of their reciprocals and as the squares of the variables). So for now, let $a, b, c > 0$.

By the geometric-harmonic means inequality, we have
$$abc \geq \left( \frac{3}{a^{-1} + b^{-1} + c^{-1}} \right)^3 \geq 1.$$
The arithmetic-geometric means inequality gives us
$$a^2 + 4b^2 + 4c^2 \geq 9 \cdot \sqrt[3]{a^2 b^8 c^8}, \quad b^2 + 4c^2 + 4a^2 \geq 9 \cdot \sqrt[3]{a^8 b^2 c^8}, \quad \text{and} \quad c^2 + 4a^2 + 4b^2 \geq 9 \cdot \sqrt[3]{a^8 b^8 c^2}.$$
From this, we obtain
$$\begin{aligned} (a^2 + 4(b^2 + c^2))(b^2 + 4(c^2 + a^2))(c^2 + 4(a^2 + b^2)) &\geq 729 \cdot \sqrt[3]{a^{18}b^{18}c^{18}} \\ &= 729 \cdot (abc)^2 \\ &\geq 729. \end{aligned}$$
Since equality holds for $a = b = c = 1$, we see that the maximum $m$ we are searching for is equal to $729$. Equality holds if the absolute values of all variables are equal to $1$, and we therefore have eight possible triples of variables for which equality holds, namely $(a, b, c) = (\pm 1, \pm 1, \pm 1)$.

$\square$