The same formula holds for $\chi(X^{[n]})$ of any topological space for which $\chi(X)$ is defined and behaves in the expected way for unions, Cartesian products, and quotients by a finite free action. (This includes the category of all algebraic varieties over $\mathbb C$.) That is because the proof only uses additivity and multiplicativity of Euler characteristic, and the behavious under quotient by free actions, and nothing elseC$.

For example, $$\chi(X^{[2]})= \frac{\chi(X\times X)- \chi(\operatorname{diag} X)}{2} + \chi(X) = \frac{ \chi(X)^2 + \chi(X) }{2},$$ etc. No matter what whether $X$ is a smooth curve or a singular 3-fold.

The same formula holds for $\chi(X^{[n]})$ of any topological space for which $\chi(X)$ is defined. (This includes all algebraic varieties over $\mathbb C$.) That is because the proof only uses additivity and multiplicativity of Euler characteristic, and the behavious under quotient by free actions, and nothing else.

For example, $$\chi(X^{[2]})= \frac{\chi(X\times X)- \chi(\operatorname{diag} X)}{2} + \chi(X) = \frac{ \chi(X)^2 + \chi(X) }{2},$$ etc. No matter what $X$ is.