Let p(x) = f(x - 4). Then |p(±1)| = |p(±2)| = |p(±3)| = 12. Supposing p is odd, we want to find a & c such that p(x) = ax^3 + cx and |p(1)| = |p(2)| = |p(3)| = 12. We can turn this into a linear system of equations -- namely

- a + c = p(1) = ±12
- 8a + 2c = p(2) = ±12
- 27a + 3c = p(3) = ±12

Because the above linear system is overdetermined, we should be able to express one of the three left-hand sides as a linear combination of the other two. Indeed, 4(8a + 2c) - 5(a + c) = 27a + 3c, so we immediately see that p(1) & p(2) must have the same sign. Which sign p(1) & p(2) share doesn't matter, because swapping the sign would amount to negating p and we're only interested in the absolute value of f(0) = p(-4). WLOG we'll therefore solve

- a + c = -12
- 8a + 2c = -12

Skipping the boring details, a = 2 & c = -14, so p(x) = 2x^3 - 14x and |p(-4)| = |2*64 - 56| = 72.

Anyone else care to prove that the p I've found is the only solution (up to sign) to the system |p(±1)| = |p(±2)| = |p(±3)| = 12?