Does the pattern of identities for $(x+y+z)^{2n-1}$ continue past $2 n - 1 = 5$? We have the following identities.
\begin{align}
(x+y+z)^3&=x^3+y^3+z^3+3(x+y)(y+z)(z+x)\\
(x+y+z)^5&=x^5+y^5+z^5+5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)
\end{align}
I'm wondering is there any pattern like above in the expansion of $(x+y+z)^{2n-1}$ where $n\in\mathbb{Z}^+$? The pattern I see so far is appearance of the coefficient $2n-1$ and $(x+y)(y+z)(z+x)$.

The above expansions can be found by factoring $(x+y+z)^{t}-x^{t}-y^{t}-z^{t}$ by using Cyclic Polynomials. For example for $t=5$ the factorization is provided on Brilliant.

I tried factoring $(x+y+z)^{7}-x^{7}-y^{7}-z^{7}$ using the method I got:
$$(x+y)(y+z)(z+x) [A(x^4+y^4+z^4)+B(x^3y+y^3z+z^3x)+C(x^2y^2+y^2z^2+z^2x^2)]$$
Here I'm not sure whether I should include $D(xy^3+yz^3+zx^3)$ too or not.

Yes you do keep the factors $x+y,x+z,y+z$. We can prove that, for instance, $x+z$ divides $(x^k+y^k+z^k)-(x+y+z)^k$ for all odd $k$ by observing that the difference goes to zero simultaneously with the claimed factor $x+z$ when we specify $z=-x$.
