All I have are three small observations.

For all $a,c\in \mathbb N^+$, $(2a,1,2c)$ is a second player win. Player two's winning strategy is to divide the board into adjacent pairs, and to respond with the square paired with player one's last move.

I tried to find a similar pairing strategy for a $(2a,2b,2c)$ board, but failed. In the game where only orthogonal rows count as a win, there is a sort of pairing strategy for the second player: divide the board into $2\times 2$ blocks, and make a winning move if it exists, otherwise respond diagonally opposite in the same block. Unfortunately, this fails when diagonals are wins.

The game $(a,1,3)$ is the octal game 0.11337, which is equivalent to 0.007 with an offset, as Timothy Chow said. That is, when $n\ge 2$, an empty $(n,1,3)$-board of length $n$ is equivalent to a 0.11337-heap of size $n$, which it turn equivalent to a 0.007-heap of size $n-2$ .

All I have are three small observations.

For all $a,c\in \mathbb N^+$, $(2a,1,2c)$ is a second player win. Player two's winning strategy is to divide the board into adjacent pairs, and to respond with the square paired with player one's last move.

I tried to find a similar pairing strategy for a $(2a,2b,2c)$ board, but failed. In the game where only orthogonal rows count as a win, there is a sort of pairing strategy for the second player: divide the board into $2\times 2$ blocks, and make a winning move if it exists, otherwise respond diagonally opposite in the same block. Unfortunately, this fails when diagonals are wins.

The game $(a,1,3)$ is the octal game 0.11337, which is equivalent to 0.007 with an offset, as Timothy Chow said. That is, when $n\ge 2$, an empty $(n,1,3)$-board is equivalent to a 0.11337-heap of size $n$, which it turn equivalent to a 0.007-heap of size $n-2$ .