A Sudden Dip

Notice that $f(x) = x$ holds for $x = 1, 2, \dots, 2016.$ Thus, we can write $f(x)$ as

$f(x) = x + g(x) \prod_{i = 1}^{2016} (x - i),$

for some polynomial $g(x).$ Substituting in $x = 2017$ gives $f(2017) = 2017 + g(2017)(2016!).$ Since $f(2017) = 1,$ we set $2017 + g(2017)(2016!) = 1$ to get $g(2017) = -\frac{1}{2015!}.$

Now, notice that we want the degree of $f$ to be as small as possible. Since $\deg \left ( \prod_{i = 1}^{2016} (x - i) \right ) > \deg(x),$ we see that in order to get a minimal degree of $f,$ we must minimize the degree of $g.$ Thus, we must have $\deg g = 0,$ meaning that $g$ is a constant function. We already found $g(2017) = -\frac{1}{2015!},$ therefore $g(x) = -\frac{1}{2015!}.$

Our final polynomial is

$f(x) = x - \dfrac{1}{2015!} \prod_{i = 1}^{2016} (x - i).$

Substituting in $x = 2018$ yields $f(2018) = 2018 - \frac{1}{2015!}(2017!) = 2018 - 2017(2016),$ whose absolute value has last three digits $17(16) - 18 = \boxed{254}.$