Consider the diagonals going northwest/southeast. (I'll call these primary diagonals and the ones going the other way secondart diagonals.) Look at the top-right-most primary diagonal that contains any alive squares. None of those squares can survive, since they can have at most one cell alive in their neighborhood. This means that in each step with living cells, a primary diagonal that contains at least one cell will be fully killed off. (1)
Can we generate cells fast enough to keep them alive? To generate new live cells, we need two adjacent live cells on the same primary diagonal. There is a top-left-most cell in the top-right-most primary diagonal that can only be used to generate one new cell - that means not every cell can be used twice, so we will generate at most $n-1$ cells (where $n$ is the number of cells on "death row", so to speak).
We can't space them out, because then the diagonal below the space will die off too. Call a "group" a set of cells that, if left to fully generate without applying the death rule, would be connected horizontally, vertically, and primary diagonally. Splitting the live cells into two groups only makes them die faster, since both groups' topmost primary diagonal would die off. The optimal strategy to keep cells alive would be to have them all in one group.
Let an "oriented triangle of size $n$" be a set of cells (alive or dead) forming a shape like this:
There are $n$ cells along the bottom, left, and along the primary diagonal.
Since all the live cells are grouped together, there is an oriented triangle bounding them of size at most $2016$.
In each step, the biggest primary diagonal of the oriented triangle will die off (by (1)).
Therefore, in each step, the size of the oriented triangle containing the group will decrease by at least $1$.
Therefore all the cells will die in at most $2016$ steps. $\blacksquare$