Nice question!

For the maximum number of pairwise non-defending rooks,
Will Sawin proved an upper bound of $(2n/3) + 1$
in his comment to the original question. This bound is attained,
at least to within $O(1)$, by two rows of $n/3 - O(1)$ rooks each,
starting from around $(2n/3,n/3,0)$ and $(n/3,2n/3,1)$
and proceeding by steps of $(-1,-1,2)$ until reaching the
$y=0$ or $x=0$ edge of the triangle. This construction
generalizes Sawin's five-Rook placement for $n=6$.

On further thought, it seems we actually achieve
$\lfloor (2n/3) + 1 \rfloor$ exactly for all $n$.
Here's how it works for $n=12$ and $n=15$,
with $(2n/3)+1 = 9$ and $11$ respectively:

```
.
. .
. . .
. . . . .
. . . . . . .
. . . R . . . . .
. . . . . . . . . . R
R . . . . . R . . . . . .
. . . . . R . . . . . . . R .
. R . . . . . . . R . . . . . . .
. . . . . . R . . . . . . . . . R . .
. . R . . . . . . . . . R . . . . . . . .
. . . . . . . R . . . . . . . . . . . R . . .
. . . R . . . . . . . . . . . R . . . . . . . . .
. . . . . . . . R . . . . . . . . . . . . . R . . . .
. . . . R . . . . . . . . . . . . . R . . . . . . . . . .
```

Starting from such a solution with $n=3m$, we can add an empty row
to get an optimal solution for $n=3m+1$, and remove an edge
(and the Rook it contains) to get an optimal solution for $n=3m-1$.
So this should solve the problem for all $n$.

Jeremy Martin also asks:

More generally, is anything known about the graph whose vertices
are these ordered triples and whose edges are rook moves?

I don't remember reading about this graph before.
Experimentally (for $3 \leq n \leq 16$) its adjacency matrix
has all eigenvalues integral, the smallest being $-3$ with huge multiplicity
$n-1\choose 2$; more precisely:

**Conjecture.** For $n \geq 3$ the eigenvalues of the adjacency matrix are:
a simple eigenvalue at the graph degree $2n$; a $n-1\choose 2$-fold
eigenvalue at $-3$; and a triple eigenvalue at each integer
$\lambda \in [-2,n-2]$, except that $\mu := \lfloor n/2 \rfloor - 2$
is omitted, and $\mu - (-1)^n$ has multiplicity only $2$.

This is probably not too hard to show. For example, the $\lambda = -3$
eigenvectors constitute the codimension-$3n$ space of functions
whose sum over each of the $3(n+1)$ Rook lines vanishes.
*[Added later: in the comment Jeremy Martin reports that
he and Jennifer Wagner already made* **and proved** *the same conjecture.]*

Given that the minimal eigenvalue is $-3$,
it follows by a standard argument in "spectral graph theory"
that the maximal cocliques have size at most $3(n+1)(n+2)/(4n+6) = 3n/4 + O(1)$.
But that's asymptotically worse than $2n/3 + O(1)$, though it's still
good enough to prove the optimality of Will Sawin's cocliques of size
$5$ for $n=6$ and of size $7$ for $n=9$.

Here's some **gp** code to play with this graph and its spectrum:

```
{
R(n)=
l = [];
for(a=0,n,for(b=0,n-a,l=concat(l,[[a,b,n-a-b]])));
matrix(#l,#l,i,j,vecmin(abs(l[i]-l[j]))==0) - 1
}
```

running "R($n$)" puts a list of the vertices in "l" and returns
the adjacency matrix with the corresponding labeling. So for instance

```
matkerint(R(7)-2)~
matkerint(R(8)-1)~
```

returns matrices whose rows are nice generators of the
$2$-dimensional eigenspaces of the $n=7$ and $n=8$ graphs.