question: I am looking for the literature with the result or the computation of Pin- bordism group: $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$. Can someone point out some useful ways to do this or any helpful Refs?

Some helpful background: There is isomorphism between the following Spin and the Pin- bordism group, known as the Smith isomorphism: $$ \Omega^{Spin}_{d+1}(B\mathbb{Z}_2)' \to \Omega^{Pin-}_{d}(pt) $$ in particular, the $\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)'$ is not exactly the the usual Spin bordism group $\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)'$, but the reduction, based on a relation: $$ \Omega^{Spin}_{d+1}(BG)=\Omega^{Spin}_{d+1}(BG)' \oplus \Omega^{Spin}_{d+1}(pt) $$ where the reduction of the spin bordism group $\Omega^{Spin}_{d+1}(BG)$ to $\Omega^{Spin}_{d+1}(BG)'$ gets rid of the $\Omega^{Spin}_{d+1}(pt)$. This part has something to do with the kernel of the forgetful map to $\Omega^{Spin}_{d+1}(pt)$.

In principle, to compute $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$, we may prove and use the following relations (any comments about this approach):

$$ \Omega^{Spin}_{d+1}(B(\mathbb{Z}_2)^2)' \to \Omega^{Pin-}_{d}(B\mathbb{Z}_2)? $$

Some useful info:

$\Omega^{Pin-}_1(pt)=\mathbb{Z}_2, \Omega^{Pin-}_2(pt)=\mathbb{Z}_8, \Omega^{Pin-}_3(pt)=0, \Omega^{Pin-}_4(pt)=0$

$\Omega^{Spin}_1(B\mathbb{Z}_2)=\mathbb{Z}_2^2, \Omega^{Spin}_2(B\mathbb{Z}_2)=\mathbb{Z}_2^2, \Omega^{Spin}_3(B\mathbb{Z}_2)=\mathbb{Z}_8, \Omega^{Spin}_4(B\mathbb{Z}_2)=\mathbb{Z}$

This is the reference that I have at hand: Kirby-Taylor, Pin structures on low-dimensional manifolds

I am willing to hear some guidance along this line of thinking, or related issue.