Since $ f(x) $ has integer coefficients, the Integer Root Theorem tells us that all integer roots of $ f(x) $ must divide the constant term $ 66=2\cdot 3\cdot 11 $. Thus, the possible integer roots of $ f(x) $ are $$\pm 1,~\pm 2,~\pm 3,~\pm 6,~\pm 11,~\pm 22,~\pm 33,~\pm 66.$$Moreover, since we know that all roots of $ f(x) $ are integers, we know that all roots of $ f(x) $ appear in the list above. Now we apply Vieta's formulas. The product of the roots of $ f(x) $ is $ (-1)^n\cdot\frac{a_0}{a_n} $, which is $ 33 $ or $ -33 $. Also, the sum of the roots is $ -\frac{a_{n-1}}{a_n}=-\frac{a_{n-1}}2 $. Thus, in order to minimize $ |a_{n-1}| $, we should make the absolute value of the sum of the roots as small as possible, working under the constraint that the product of the roots must be $ 33 $ or $ -33 $. We now consider two cases. Case 1 is that one of $ 33,-33 $ is a root, in which case the only other possible roots are $ \pm 1 $. In this case, the absolute value of the sum of the roots is at least $ 32 $. The alternative, Case 2, is that one of $ 11,-11 $ is a root and one of $ 3,-3 $ is a root. Again, the only other possible roots are $ \pm 1 $, so the absolute value of the sum of the roots is at least $ 11-3-1=7 $, which is better than the result of Case 1. If the absolute value of the sum of the roots is $ 7 $, then $ |a_{n-1}|=7|a_n|=7\cdot 2=14 $. Therefore, we have shown that $ |a_{n-1}|\ge 14 $, and we can check that equality is achieved by \begin{align*} f(x) &= 2(x+11)(x-3)(x-1) \\ &= 2x^3+14x^2-82x+66, \end{align*}which has integer coefficients and integer roots. So the least possible value of $ |a_{n-1}| $ is $ \boxed{14} $.