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][1], or Theorem 9.1.1 of the book [Unstable Modules over the Steenrod Algebra and Sullivan's Fixed Point Conjecture][2].  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.)

  [1]: https://www.math.uchicago.edu/~may/BOOKS/alaska.pdf
  [2]: https://press.uchicago.edu/ucp/books/book/chicago/U/bo3634919.html