$ f(0) = c$ $ f(\frac {1}{2}) = \frac {a}{4} + \frac {b}{2} + c$ $ f(1) = a + b + c$. solving this system for $ (a,b,c)$,we have $ (a,b,c) = (2f(0) - 4f(\frac {1}{2}) + 2f(1), - 3f(0) + 4f(\frac {1}{2}) - f(1),f(0))$. thus by the triangle inequality we have $ |a| + |b| + |c| = |2f(0) - 4f(\frac {1}{2}) + 2f(1)| + | - 3f(0) + 4f(\frac {1}{2}) - f(1)| + |f(0)| \le |2f(0)| + |4f(\frac {1}{2})| + |2f(1)| + | - 3f(0)| + |4f(\frac {1}{2})| + | - f(1)| + |f(0)| \le 2 + 4 + 2 + 3 + 4 + 1 + 1 = 17$. this equality holds if and only if $ f(0) = f(1) = \pm1,f(\frac {1}{2}) = \mp1$ i.e. $ f(x) = \pm(8x^2 - 8x + 1)$ ,which surely satisfy the condition.So finally we see the greatest value is $ \boxed{17}$.