Let $XY = qYX$ and $YZ = q ZY$ and $ZX = qX Z$ for $q^n = 1$ a root of unity. Does it follow that:
$$(X+Y+Z)^n = X^n + Y^n + Z^n $$
In other words, all the middle terms cancel out. If $q^m \neq 1$ for $0 < m < n$ (that is a primtive root of unity) at least we could group into monomial terms.
$$ \sum_{a+b+c = n} \Big[ \sum q^{\dots }\Big] X^a Y^b Z^b $$
After I'm not 100% sure the middle terms all cancel out. Since neither the left and right sides change under the map $q \mapsto q^k$ (for $k \neq 1$) the only possible invariants are constants... so these coefficients certainly do not depend on $q$ but they might be zero.
If $q=1$ the middle numberes are the multinomial coefficients:
$$ (X+Y+Z)^n = \sum_{a+b+c = n} \binom{n}{a,b,c} X^a Y^b Z^b $$
Let $n = p$ a prime number if it makes this easier, so that $q = e^{\frac{2\pi i }{p}}$. If $p = 5$ we'd have something like:
$$ (X+Y+Z)^5 = X^5 + Y^5 + Z^5 $$
Just to answer the original question (with $ZX = qXZ$): the answer is "No". Here is a counterexample for $n = 3$, computed using SageMath (containing a lot of my old legacy code):
The answer is "False", which signifies that $\left(X+Y+Z\right)^3 - \left(X^3 + Y^3 + Z^3\right)$ is not in the ideal generated by the relations you gave.
Alternatively, you can check this by using the multigrading on the free algebra (in which the degree of a word $w$ is $\left(\text{number of $X$'s in $w$}, \text{number of $Y$'s in $w$}, \text{number of $Z$'s in $w$}\right)$). Clearly, if $\left(X+Y+Z\right)^3 - \left(X^3 + Y^3 + Z^3\right)$ would lie in the ideal, then the multidegree-$\left(1,1,1\right)$ component $XYZ+YZX+ZXY+XZY+YXZ+ZYX$ of $\left(X+Y+Z\right)^3 - \left(X^3 + Y^3 + Z^3\right)$ would also lie in this ideal (because the ideal is homogeneous with respect to this multigrading). But an easy verification shows that this is not the case.