Define $M(a, b, c) = \max_{x \in [-1, 1]} |x^3 + a x^2 + b x + c|$. Observe that
$$\max_{x \in [-1, 0]} |x^3 + a x^2 + b x + c| = \max_{x \in [0, 1]} |-x^3 + a x^2 - b x + c|.$$
Since $\max(|\alpha + \beta|, |\alpha - \beta|) = |\alpha| + |\beta|$, then
$$M(a, b, c) = \max_{x \in [0, 1]} (|x^3 + b x| + |a x^2 + c|) \geq \max_{x \in [0, 1]} |x^3 + b x| = M(0, b, 0).$$
Now we can put $x = \cos \theta$ in $x^3 + b x$, where $\theta \in [0, \frac{\pi}{2}]$. Then $x^3 + b x = \cos^3 \theta + b \cos \theta = \frac{1}{4}(\cos(3\theta) + 3 \cos \theta) + b \cos \theta = \frac{1}{4}(\cos(3\theta) + (3 + 4b) \cos \theta)$.
Define $m_\alpha = \max_{\theta \in [0, \frac{\pi}{2}]} |\cos(3\theta) + \alpha \cos \theta|$.
Using the values at $\theta = 0$ and $\theta = \pi/3$, gives $m_\alpha \ge \max(|1 + \alpha|, |1 - \alpha/2|) \ge 1 + |\alpha|/2 \ge 1$, all $\alpha$. Clearly $m_0 = 1$, so $\min_\alpha m_\alpha = 1$ and this occurs if and only if $\alpha = 0$, that is $b = -\frac{3}{4}$.
So $\min_b M(0, b, 0) = M(0, -\frac{3}{4}, 0) = \frac{1}{4}$.
So the minimum is $\frac{1}{4}$, which occurs for $x^3 - \frac{3}{4}x$.