Given a polynomial $f(z)=a_0z^n+a_1z^{n-1}+\cdots+a_n$ we certainly have $\max_{|z|\leq1}|f(z)|\leq |a_0|+|a_1|+\cdots+|a_n|.\tag{1}$ More generally, analytic functions with $\ell^1$ coefficients are bounded in the unit disc and the supremum is bounded by the $\ell^1$-norm - that is if $f(z)=\sum_na_nz^n$ and $\|f\|_{\ell^1}=\sum|a_n|<\infty$ then $f$ is bounded in the unit disc (in fact it belongs to the so called disc algebra) and $\sup_{|z|<1}|f(z)|\leq\|f\|_{\ell^1}$
In our case, $f(z)=2z^2-1$ it is rather easy to guess a point $z_0$ (or two) where $|f(z_0)|=3$ and hence we have equality in (1).
However, in general we cannot get equality in (1), which is easiest realized by an example: Consider $g(z)=(z+1)(z^2-1)=z^3+z^2-z-1$ then $\|g\|_{\ell^1}=4$ - now if we would have equality in (1) then $4=\max|g|\leq\max|z+1|\cdot|z^2-1|=2\cdot2$ This means that the max must be attained at $z=1$ where $|z+1|$ is maximized, but this is not the case (since the other factor is 0 there).