[UPDATE: Connor Malin's answer is definitely better than mine, but I will leave mine here in case the approach turns out to be useful for some other purpose.]
Put $X=\operatorname{Conf}_n(M)$ and $Y=\operatorname{Conf}_2(M)$. We want to show that $X^{h\Sigma_n}$ is empty. There is an evident projection $X\to Y$ which is equivariant for the evident copy of $\Sigma_2$ in $\Sigma_n$ so we get a projection $X^{h\Sigma_n}\to Y^{h\Sigma_2}$, so it will suffice to show that $Y^{h\Sigma_2}=\emptyset$. I think that this holds provided that $Y$ has the $\Sigma_2$-equivariant homotopy type of a finite $\Sigma_2$-CW complex, which I think should be true under mild conditions on $M$ (despite the fact that $Y$ is certainly not compact).
To see this, let $Y_2$ be the $2$-adic completion of $Y$ in the sense of Bousfield and Kan. This is obtained by applying a functorial simplicial construction to the singular complex $SY$, so we have maps $Y\xleftarrow{p}|SY|\xrightarrow{q}Y_2$ in which $p$ is a weak equivalence, and everything is $\Sigma_2$-equivariant. If we can show that $Y_2^{h\Sigma_2}=\emptyset$ then it will follow that $|SY|^{h\Sigma_2}=\emptyset$, and the homotopy fixed point functor preserves weak equivalences, so $Y^{h\Sigma_2}=\emptyset$.
We can now apply the Sullivan Conjecture, in the form proved by Lannes: see Theorem VIII.1.2 of the Alaska book, or Theorem 9.1.1 of the book Unstable Modules over the Steenrod Algebra and Sullivan's Fixed Point Conjecture. Assuming the stated finiteness condition, that says that the natural map $(Y^{\Sigma_2})_2\to Y_2^{h\Sigma_2}$ is a weak equivalence, but here $Y^{\Sigma_2}=\emptyset$, so $Y_2^{h\Sigma_2}=\emptyset$ as required. (Some versions of the Sullivan Conjecture require that $Y$ should be a nilpotent space, but the version proved by Lannes does not.)