If $M$ is a connected smooth manifold, then it is easy to show that there is a sequence of connected compact smooth submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=1}^\infty(M_i)^\circ$.
I would guess it should also be true that if $M$ is a connected topological manifold then there is a sequence of locally tame connected compact submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=1}^\infty(M_i)^\circ$. How would one try to prove such a statement? The only proof I know of the statement in the smooth category is to start with any exhaustion by open sets with compact closure and then "smooth" their boundaries. However, modifying an open set in a topological manifold so that its boundary is a tamely embedded codimension 1 submanifold seems much more delicate (and perhaps there is even an obstruction to doing it!).
Best Answer
Since topological manifolds of dimension $\le 3$ are smoothable, the question is about manifolds of dimension $\ge 4$. Kirby and Siebenmann proved for $n\ge 6$ that every topological $n$-manifold admits a handle decomposition; this was extended to $n=5$ by Freedman and Quinn (I think, it is Quinn's paper "Ends of maps, III"). This applies to noncompact manifolds as well. Using this handle decomposition you can easily construct the required exhaustion (just use finitely many handles). This settles the problem in all dimensions but 4.
Handle decomposition is known to fail in dimension 4, but there is an alternative argument: Take $N^5=M^4\times R$, construct an exhaustion of $N$ as above by compact submanifolds $S_i$. Now, Quinn proved in 1988 a topological transversality theorem in all dimensions ("Topological transversality holds in all dimensions"), which allows you to perturb each $S_i$ to $S_i'$ whose boundary is transversal to $M\times 0$. Then $S_i'\cap M$ will be the required exhaustion.