jmc

Each of the six values $f(1),$ $f(2),$ $f(3),$ $f(5),$ $f(6),$ $f(7)$ is equal to 12 or $-12.$ The equation $f(x)=12$ has at most three roots, and the equation $f(x)=-12$ has at most three roots, so exactly three of the values are equal to 12, and the other three are equal to $-12.$

Furthermore, let $s$ be the sum of the $x$ that such that $f(x)=12.$ Then by Vieta's formulas, the sum of the $x$ such that $f(x)=-12$ is also equal to $s.$ (The polynomials $f(x)-12$ and $f(x)+12$ only differ in the constant term.) Hence, $$2s=1+2+3+5+6+7=24,$$so $s=12.$

The only ways to get three numbers from $\{1,2,3,5,6,7\}$ to add up to 12 are $1+5+6$ and $2+3+7.$ Without loss of generality, assume that $f(1)=f(5)=f(6)=-12$ and $f(2)=f(3)=f(7)=12.$

Let $g(x)=f(x)+12.$ Then $g(x)$ is a cubic polynomial, and $g(1)=g(5)=g(6)=0,$ so $$g(x)=c(x-1)(x-5)(x-6)$$for some constant $c.$ Also, $g(2)=24,$ so $$24=c(2-1)(2-5)(2-6).$$This leads to $c=2.$ Then $g(x)=2(x-1)(x-5)(x-6),$ so $$f(x)=2(x-1)(x-5)(x-6)-12.$$In particular, $|f(0)|=\boxed{72}.$