I am not sure whether this answers your question in full, but I can suggest an idea. It is based on triangulated, oriented, labelled manifolds. You can drop the orientation and then move from coefficients in $\bf Z$ to $\bf Z/2Z$.
Define $\Omega_0^{tol}$ to be the group of cobordism classes of two-coloured points. The superscript $tol$ stands for triangulated, oriented, labelled. Using the notation from your question, $\Omega_0^{tol} = \Omega_0^{SO} = \bf Z$.
Moving on to 1-dimensional closed manifolds. Here, these are oriented, 3-coloured circles.
The key ingredient to determine $\Omega_1^{tol}$ is the generalized Lemma of Sperner ini combination with a degree function on the labelling.
It says: If you triangulate a 2-disk and label it with 3 colours, such that the labelling or colouring of the boundary circle has degree $k$, then
$$\sum \Delta^+ - \sum\Delta^- = k$$
where $\Delta^+$ are 3-coloured triangles with one orientation and $\Delta^-$ are 3-coloured triangles with the opposite orientation.
My illustration at the top shows an example for $k = 2$ where both triangles have the same orientation. You can see the details about degree and orientation in the reference below. It basically works like this: the degree of the labelling or colouring is defined through counting the colour changes e.g. from blue to yellow on the boundary circle $M$. I have put $+1$ and $+1$ in the diagram to show that count.
Now here is the key idea. Translate that into cobordism. See the 3-coloured triangles as holes, and see all other triangles as 2-dimensional, filled triangles. Saying it differently, regard the 3-coloured triangles as circles $S^1$ while the other triangles form a triangulated 2-manifold $X$. To illustrate this idea, the 3-coloured triangles $N_1$ and $N_2$ in my picture are not filled while the interiors of all other triangles are shaded (with the meaning they are subsets of a 2-manifold; their union defines a 2-manifold).
So, translated to cobordism, the generalized Lemma of Sperner asserts: for each oriented, 3-coloured circle $M$ with labelling degree $k$ you can find a 2-manifold $X$ (triangulated, oriented, 3-coloured disk with holes) such that
$$\partial X = M \sqcup -N_1 \sqcup -N_2 \sqcup \ldots \sqcup -N_k$$
This means $M \cong N_1 \sqcup N_2 \sqcup \ldots \sqcup N_k$ and you have a cobordism equivalence relation. $-N_1$ is the inverse of $N_1$ which just means two vertex colours of the triangle are flipped (i.e. orientation is reversed).
Importantly, we need to be careful with the orientations here. If you start with an orientatin for $M$, the orienation of $X$ and $N_i$ have to be compatible (such that the orientations of $M$ and $N_i$ can be seen as induced by the orientation of $X$).
The degree $k$ of the labelling can take any value $k\in \bf Z$, and indeed $\Omega_1^{tol}$ is generated by a 3-coloured triangle with degree $+1$. In summary
$$\Omega_1^{tol}={\bf Z} \neq \Omega_1^{SO}$$
so this really defines a cobordism theory that is different from the oriented smooth one.
The generalized Lemma of Sperner also holds for higher dimensions with the same formula for simplices (and of course $d+1$ colours for dimension $d$). Hence, if you restrict yourself to spheres $S^d$, you will get $\Omega_d^{tol}=\bf Z$ (but I am not sure this restriction is meaningful). If you can show that the generalized Lemma of Sperner (or variants of it) also holds for the torus, the n-holed torus etc, you can determine $\Omega_2^{tol}$ for surfaces and move on to higher dimensional manifolds.
Regarding degree of labelling and generalized Lemma of Sperner, here is the reference: Michael Prüfer and Hans Willi Siegberg (1981): Complementary Pivoting and the Hopf Degree Theorem. Journal of Mathematical Analysis and Applications, Volume 84, Issue 1, November 1981, Pages 133-149.
