You're right. My manifold is only almost hyperkahler. I edited the question, but this makes it less likely that someone has seen this manifold "in the wild".

It would be nice if there was a construction of this manifold that didn't require knowing a priori that $bP_8 = \mathbb{Z}/28$. I was hoping to use this manifold to prove that $240 \sigma = 0$, but since the computation that $bP_8 = \mathbb{Z}/28$ presupposes $\pi_7^{st} = \mathbb{Z}/240$, this is circular reasoning.

That is a great way to see that $p_1$ is trivial! Thanks @TomMrowka btw it was your beautiful answer on the linked MO question that got me thinking along these lines. I will accept this answer, but I still think it is 7 copies of $S^4 \times S^4$ that are needed. If I am doing things right the first half of the construction (the 28 Milnor plumbings + cap) has $\chi = 8 *28 + 2 = 226$ while the connect sum of 8 $S^4 \times S^4$s has $\chi = 8 *2 +2 = 18$. The connect sum of these two parts then has $\chi(M) = 226 + 18 - 2 = 242$. If you use 7 $S^4 \times S^4$s, you get 240 instead.

The Barwick-Kan model structure on relative categories is lifted from bisimplicial sets using the functor $N_\xi$. So $N_\xi$ (and hence $N$) is automatically a relative functor (aka homotopical). It preserves all weak equivalences (not just between fibrant objects). In fact that is how the weak equivalences between relative categories are defined.

( I meant $L$ is a bundle on $\mathbb{OP}^1$, not $\mathbb{OP}^2$)

Then if we blow up those 256 points using $Y$, we get a manifold $M$ with $\chi(M) = 256 y - 128$. But then 240 is not an integer solution to this equation. So any Kummer-esque construction is going to have to be more involved.

@DylanWilson I thought about that a little. If we just proceed naively and take the abelian group $T^8$ modulo $a \mapsto -a$ then we get a singular (orbifold) quotient with $2^8 = 256$ singular points. If we cut out balls in a neighborhood of each of these points we get a space $W$ with $\chi(W) = -256/2 = -128$ and $\partial W =$ 256 copies of $\mathbb{RP}^7$. I am not sure how to resolve the singularities. We must choose a manifold $Y$ with boundary $\mathbb{RP}^7$. Let $y = \chi(Y)$ be its Euler characteristic. (cont)

Also for others who might read this and be confused like I was, one key point is that for this bundle L on $\mathbb{OP}^2$, we have $e(L)$ is the generator but $p_2(L) = 6 e(L)$. This follows from the calculation that the map $S^8 \to BSO$ which is the generator in $\pi_8$ induces multiplication by 6 in cohomology (generated by $p_2$ on the RHS). This is also why on String manifolds there is a $p_2/6$ class.

This is really really great, thank you! I have lots of questions. When you say the Milnor pluming, I take it you mean the pluming on the $E_8$-graph, yes? I am confused about two points. The first is: did you really mean 8 copies of $S^4 \times S^4$? or did you mean 7 copies? When I tried to compute the Euler characteristic of M I am off by 2, which could mean I made an arithmetic mistake, or that we really want 7 copies. More importantly, I see how the argument works if we know that $p_1(TM) = 0$. Can you remind me how we can see that?

The Octionions $\mathbb{O}$, viewed as an 8-manifold, admit the kind of structure I have in mind, as does the 8-torus $T^8 = \mathbb{O}/ \Lambda$, or any (unstably) framed 8-manifold. Also the tautological rank eight bundle on $\mathbb{OP}^1 = S^8$ is also a vector bundle with this sort of structure. They are associated to "$GL_1(\mathbb{O})$-principal bundles" where $GL_1(\mathbb{O})$ is a "non-associative Lie group". This makes sense since each of these bundles admits a trivialization over a cover with no non-trivial triple intersections (so associativity doesn't factor into it).

Thanks @JesperGrodal, does this mean that this category SH/acyclic is blind to $v_n$-periodic phenomena?

What is the easiest example of an acyclic spectrum? I thought there is a stable Whitehead theorem which says that all such spectra are contractible? Maybe that is just for connective spectra? or I am being stupid?