jmc

We have that $$(n+i{)}^2=n^2+2ni+i^2=(n^2-1)+(2n)i,$$and $$(n+i{)}^3=n^3+3n^{2}i+3n{i}^2+i^3=(n^3-3n)+(3n^2-1)i.$$By the Shoelace Theorem, area of the triangle with vertices $(n,1),$ $(n^2-1,2n),$ and $(n^3-3n,3n^2-1)$ is $$\begin{array}{rl}& \frac{1}{2}\left|(n)(2n)+(n^2-1)(3n^2-1)+(n^3-3n)(1)-(1)(n^2-1)-(2n)(n^3-3n)-(3n^2-1)(n)\right|\\ & =\frac{1}{2}(n^4-2n^3+3n^2-2n+2)=\frac{1}{2}[(n^2-n+1{)}^2+1].\end{array}$$Thus, we want $n$ to satisfy $$\frac{1}{2}[(n^2-n+1{)}^2+1]>2015,$$or $(n^2-n+1{)}^2>4029.$ Checking small values, we find the smallest positive integer $n$ that works is $\boxed{9}.$