Every closed oriented seven-manifold bounds

Theorem. For every closed smooth oriented seven-manifold $M$, there exists a compact smooth oriented eight-manifold $W$ and an orientation-preserving diffeomorphism $\partial W\cong M$. Equivalently, $\Omega_7^{SO}=0.$

No simple-connectivity, homology-sphere, spin, or particular tangent-bundle assumption is imposed. Disconnected closed manifolds are included.

Exact proof chain

Stiefel–Whitney numbers and unoriented bordism, Theorem N.3 and Corollary N.4, give $G=\Omega_7^{SO}\quad\text{finite of odd order}.$ These are the geometric twofold-cover and forgetful-kernel arguments, combined with rational finite generation and rank zero. They do not themselves exclude odd-primary torsion.
For every odd prime $p$, the one-group and spectrum comparison, Sections 1–9, together with the coordinate calculation, Sections 1–5, identifies the full algebra $A=[H\mathbb F_p,\Sigma^*H\mathbb F_p]$, with its coproduct and complex-line action. It proves input O of the Thom-module argument. The computation is in all degrees, and its spectrum-map comparison explicitly kills the possible inverse-limit correction.
The odd-primary Thom module, Sections 2–5, applies using that full operation-algebra computation. It gives $H^*(MSO;\mathbb F_p) \cong\bigoplus_\lambda \Sigma^{m_\lambda}(A\otimes_E\mathbb F_p),\qquad E=\Lambda(Q_0,Q_1,\ldots),$ where $A$ is free as a right $E$-module, $|Q_i|=2p^i-1,\qquad 4\mid m_\lambda.$ The module statement includes the actual action, not only the same dimension series. Its proof uses the previously supplied full odd-prime universal $BSO$ computation, the typed Whitney-sum Thom maps, and the independent distinct-root-monomial injection.
The exterior resolution, equations (1)–(9), implies $\operatorname{Ext}^{s,t}_A(H^*(MSO;\mathbb F_p),\mathbb F_p)=0 \quad\text{unless }4\mid(t-s).$ Its possible generator bidegrees obey $t-s=m_\lambda+\sum_{i\geq0}2\alpha_i(p^i-1).$ Since $p$ is odd, every summand is divisible by four, including the $i=0$ term, which is zero. In particular the entire stem $t-s=7$ is zero.
$MSO$ is connective and of finite mod-$p$ type by the Thom calculation. Finite Adams-tower detection, Sections 1–3, therefore applies with $n=7,r=1$. It gives $G=pG\qquad\text{for every odd prime }p.$ Sixteen tower stages suffice for each such assertion. This step does not replace a convergence theorem by an unexplained spectral-sequence abutment: the finite target has sixteen $H\mathbb F_p$-extension layers, and the written lifting argument gives the required divisibility.

Elimination of every primary part

If the finite odd-order group $G$ were nonzero, choose a nonzero element of order $d>1$. Some odd prime $p$ divides $d$, and its $(d/p)$-multiple has order $p$. Thus multiplication by $p$ has nonzero kernel. An endomorphism of a finite group with nonzero kernel is not onto, contradicting $G=pG$. Therefore $G=0$.

The full oriented Pontryagin–Thom correspondence in Thom spaces, transversality and the Pontryagin–Thom construction, Theorem K.1, identifies the group just calculated with geometric oriented bordism. Its construction supplies smooth collars and orientations. The zero class consequently means a compact smooth oriented filling with the asserted boundary identification, not merely a rational filling, an unoriented filling, or a filling of a multiple of $M$. This proves the theorem.

The argument also shows why the earlier finite-odd-order result was insufficient by itself: a nonzero finite group of odd order is divisible by two. The odd-prime argument is the step which excludes those remaining groups.