Let the roots of $f\left(x\right)$ be $\alpha$, $\beta$ and $\gamma$. Then $f\left(x\right)=\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)$ From $f\left(0\right)=-64$ we have $\alpha\beta\gamma=64$. Therefore $f\left(-1\right)$ $=\left(-1-\alpha\right)\left(-1-\beta\right)\left(-1-\gamma\right)$ $=-\left(\alpha +1\right)\left(\beta +1\right)\left(\gamma +1\right)$ $=-\left(65+\alpha +\beta +\gamma +\alpha\beta +\beta\gamma +\gamma\alpha\right)$ $\leq -\left(65+3\sqrt[3]{\alpha\beta\gamma}+3\sqrt[3]{\alpha^2\beta^2\gamma^2}\right)$ $=-\left(65+12+48\right)=-125$ The equality case is satisfied when $\alpha=\beta=\gamma=4$, hence the maximum value is $-125$.

What if the polynomial has only one root?

Let $f(x)=x^3+ax^2+bx+c$. $f(0)=c=-64$ $f(-1)=a-b-65$. Let $x_1,x_2,x_3$ be the roofs of $f(x)$. By Vieta we have that $x_1x_2x_3=64$ and $x_1+x_2+x_3=-a\geq12 => -12\geq a$ and $x_1x_2+x_2x_3+x_1x_3=b\geq48 => -48\geq -b$. $f(-1)=a-b-65 \leq -125$.

AmirAlison wrote: What if the polynomial has only one root? the problem itself stated that 'all roots of $f(x)$ are non-negative real numbers', so there must be 3 real roots (counted with multiplicity).

Hint

Let the roots of $f$ be $r_1,r_2,r_3$, then $f(x)=(x-r_1)(x-r_2)(x-r_3)$ with $r_1r_2r_3=64$. We'll claim $\max{(f(-1))}=-125$. It then suffices to prove \[ f(-1) \leq -125 \iff 125 \leq (1+r_1)(1+r_2)(1+r_3). \]But \[ (1+r_1)(1+r_2)(1+r_3) \geq \left(1^{\frac13}+(r_1r_2r_3)^{\frac13} \right)^3 = (1+4)^3 = 125 \]by Hölder. Thus, we're done, as the equality case can be reached for $r_1=r_2=r_3=4$. $\hfill \square$