Let $ y = \frac{1}{1 - x} $. Solving for $ x $ in terms of $ y, $ we find \[x = \frac{y - 1}{y}.\]Then \[\left( \frac{y - 1}{y} \right)^{2016} + \left( \frac{y - 1}{y} \right)^{2015} + \dots + \left( \frac{y - 1}{y} \right) + 1 = 0.\]Hence, \[(y - 1)^{2016} + y (y - 1)^{2015} + y^2 (y - 1)^{2014} + \dots + y^{2015} (y - 1) + y^{2016} = 0.\]This expands as \begin{align*} &\left( y^{2016} - 2016y^{2015} + \binom{2016}{2} y^{2014} - \dotsb \right) \\ &+ y \left( y^{2015} - 2015y^{2014} + \binom{2015}{2} y^{2013} - \dotsb \right) \\ &+ y^2 \left( y^{2014} - 2014y^{2013} + \binom{2014}{2} y^{2012} - \dotsb \right) \\ &+ \dotsb \\ &+ y^{2015} (y - 1) + y^{2016} = 0.\end{align*}The coefficient of $ y^{2016} $ is 2017. The coefficient of $ y^{2015} $ is \[-2016 - 2015 - \dots - 2 - 1 = -\frac{2016 \cdot 2017}{2} = -2033136.\]The coefficient of $ y^{2014} $ is \[\binom{2016}{2} + \binom{2015}{2} + \dots + \binom{2}{2}.\]By the Hockey Stick Identity, \[\binom{2016}{2} + \binom{2015}{2} + \dots + \binom{2}{2} = \binom{2017}{3} = 1365589680.\]The roots of the polynomial in $ y $ above are $ y_k = \frac{1}{1 - x_k} $ for $ 1 \le k \le 2016, $ so by Vieta's formulas, \[y_1 + y_2 + \dots + y_{2016} = \frac{2033136}{2017} = 1008,\]and \[y_1 y_2 + y_1 y_3 + \dots + y_{2015} y_{2016} = \frac{1365589680}{2017} = 677040.\]Therefore, \begin{align*} &\frac{1}{(1 - x_1)^2} + \frac{1}{(1 - x_2)^2} + \dots + \frac{1}{(1 - x_{2016})^2} \\ &= y_1^2 + y_2^2 + \dots + y_{2016}^2 \\ &= (y_1 + y_2 + \dots + y_{2016})^2 - 2(y_1 y_2 + y_1 y_3 + \dots + y_{2015} y_{2016}) \\ &= 1008^2 - 2 \cdot 677040 \\ &= \boxed{-338016}.\end{align*}