Answer: $M = \frac{1}{2}$.
The inequality $a^3 + b^3 + c^3 - 3abc \ge M(|a-b|^3 + |b-c|^3 + |c-a|^3)$ for $M = 1/2$ implies from
$$2(a^3 + b^3 + c^3 - 3abc) = (a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2)$$
and from obvious inequalities
$$(x + y + z)(x - y)^2 \ge |x - y|^3$$

The value $M = 1/2$ is maximal, because for very large values of $a$ and very small values of $b$, $c$, the value of $a^3 + b^3 + c^3 - 3abc$ can be very close to $a^3$, and the value of $|a-b|^3 + |b-c|^3 + |c-a|^3$ can be very close to $2a^3$.