Suppose $ m \ge 1 $. The sum of the elements of the $ m $th row is simply \[\sum_{i=0}^{m} \binom{m}{i} = \sum_{i=0}^m \binom{m}{i}1^i = (1+1)^m=2^m\] by the binomial theorem. Thus the sum of the elements of the $ k $th row, for $ k \le n $, is $ 2^k $, so the total sum of all the elements of the previous rows is \[\sum_{k=0}^{n-1} 2^k = \frac{2^n-1}{2-1} = 2^n-1\] by the formula for the sum of a geometric series. Therefore, $ f(n)=2^n-(2^n-1)=1 $ for all $ n \ge 2015 $. (That is, it is a constant!) Thus, the minimum value is clearly $ \boxed{1} $.