Your questions are answered by Hsu in his paper 4-Dimensional Topological Bordism.

In particular, associated to any closed oriented topological 4-manifold $X$ is a signature $\sigma(X) \in \Bbb Z$ and the Kirby-Seibenman class $\text{ks}(X) \in H^4(X;\Bbb Z/2) = \Bbb Z/2$. Hsu proves that these two invariants precisely classify 4-dimensional oriented topological manifolds up to bordism (and in particular $\text{ks}$ is an invariant of topological bordism).

A 4-manifold is triangulable if and only if it is smoothable. (This is true but not obvious; it invokes the 3D Poincare conjecture) If $X$ is smoothable then $\text{ks}(X) = 0$ (but the converse is not necessarily true).

The signature is an expected obstruction to finding a null-bordism, but Kirby-Seibenmann is less obvious. There is a manifold $F\Bbb{CP}^2$ ($F$ standing for Fake) which is homotopy equivalent to $\Bbb{CP}^2$ but not homeomorphic; it has $\text{ks} = 1$ and signature 1, and so is not smoothable. In particular, $F\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ has signature 0 but nonzero Kirby-Seibenmann invariant, and thus by Hsu's results it is not null-bordant.

There is a formula for spin 4-manifolds: $\text{ks}(X) = \sigma(X)/8 \pmod{2}$. In particular, $\text{ks}(X_{E8}) = \sigma(E8)/8 = 1$. Thus the $E8$-manifold is not null-bordant.

However, there are many that are: for instance, take $X_{E8} \# X_{E8}$. This is still not smoothable by Donaldson's theorem (the intersection form of a smooth 4-manifold, if positive/negative definite, is diagonalizable; $E8 \oplus E8$ is not). $\text{ks}$ is additive, and so $\text{ks}(X_{E8} \# X_{E8}) = 0$, and this manifold is null-bordant. As mentioned by Jim Conant in the comments above, because $X_{E8} \# X_{E8}$ is not triangulable, neither is the null-bordism.