jmc

Suppose $m\ge 1$. The sum of the elements of the $m$th row is simply $$\sum _{i=0}^m\left(\genfrac{}{}{0ex}{}{m}{i}\right)=\sum _{i=0}^m\left(\genfrac{}{}{0ex}{}{m}{i}\right)1^i=(1+1{)}^m=2^m$$ 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 $\boxed{1}$.