Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each has exactly three of $1,$ $2,$ $3,$ $5,$ $6,$ $7$ as roots

The two cubics differ by a constant, so the two root triples have equal sums and equal pairwise-product sums; only one split works

Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each vanishes at exactly three of $1,$ $2,$ $3,$ $5,$ $6,$ $7.$ Writing them as $c(x - r_1)(x - r_2)(x - r_3)$ and $c(x - s_1)(x - s_2)(x - s_3),$ the two cubics differ by the constant $24,$ so their $x^2$ and $x$ coefficients agree: the root triples have equal sums and equal sums of pairwise products. The only partition of $\{1,2,3,5,6,7\}$ into two triples of equal sum is $\{2,3,7\}$ and $\{1,5,6\}$ (each summing to $12$), and indeed both have pairwise-product sum $41.$

Replacing $f$ by $-f$ if necessary (which does not change $|f(0)|$), we have $f(x) = c(x-2)(x-3)(x-7)$ $+ 12$ $= c(x-1)(x-5)(x-6) - 12.$ Setting $x = 0$ gives $-42c + 12 = -30c - 12,$ so $c = 2$ and $f(0) = -42 \cdot 2 + 12 = -72.$ Thus $|f(0)| = 72.$