Since $ x^{2017} + Ax+B $ is divisible by $ (x+1)^2, $ it must have $ x=-1 $ as a root, so \[(-1)^{2017} + A(-1) + B = 0,\]or $ A=B-1 $. Then $ x^{2017} + Ax + B = x^{2017} + (B-1)x + B $. Dividing this polynomial by $ x+1 $, we have \[\begin{aligned} \frac{x^{2017} + (B-1)x + B}{x+1} &= \frac{x^{2017} + 1}{x+1} + (B-1)\\ &= (x^{2016} - x^{2015} + x^{2014} + \dots + x^2 - x + 1) + (B-1), \end{aligned}\]which must be divisible by $ x+1 $. Therefore, setting $ x=-1, $ we get \[\left((-1)^{2016} - (-1)^{2015} + (-1)^{2014} + \dots + (-1)^2 + 1\right) + (B-1) = 0,\]or $ B + 2016 = 0 $. Thus, $ B = \boxed{-2016} $.