To add to Tim Porter's excellent answer:
The story of what we now call $K_1$ of rings begins with Whitehead's work on simple homotopy equivalence, which uses what we now call the Whitehead group, a quotient of $K_1$ of the group ring of the fundamental group of a space.
On the other hand, the story of $K_0$ of rings probably begins with Grothendieck's work on generalized Riemann-Roch. What he did with algebraic vector bundles proved to be a very useful to do with other kinds of vector bundles, and with finitely generated projective modules over a ring, and with some other kinds of modules.
I don't know who it was who recognized that these two constructions deserved to be named $K_0$ and $K_1$, and viewed as two parts of something larger to be called algebraic $K$-theory. But Milnor gave the right definition of $K_2$, and Quillen and others gave various equivalent right definitions of $K_n$.
Let me try to lay out the parallels between the topological significances of Whitehead's quotient of $K_1(\mathbb ZG)$ and Wall's quotient of $K_0(\mathbb ZG)$. My main point is that both of them have their uses in both the theory of cell complexes and the theory of manifolds.
The Whitehead group of $G$ is a quotient of $K_1(\mathbb ZG)$. Its significance for cell complexes is that it detects what you might call non-obvious homotopy equivalences between finite cell complexes. An obvious way to exhibit a homotopy equivalence between finite complexes $K$ and $L$ is to by attaching a disk $D^n$ to $K$ along one half of its boundary sphere and obtain $L$. Roughly, a homotopy equivalence between finite complexes is called simple if it is homotopic to one that can be created by a finite sequence of such operations. The big theorem is that a homotopy equivalence $h:K\to L$ between finite complexes determines an element (the torsion) of the Whitehead group of $\pi_1(K)$, which is $0$ if and only if $h$ is simple, and that for any $K$ and any element of its Whitehead group there is an $(L,h:K\to L)$, unique up to simple homotopy equivalence, leading to this element in this way, and that this invariant of $h$ has various formal properties that make it convenient to compute.
One reason why you might care about the notion of simple homotopy equivalence is that for simplicial complexes it is invariant under subdivision, so that one can in fact ask whether $h$ is simple even if $K$ and $L$ are merely piecewise linear spaces, with no preferred triangulations. This means that, for example, a homotopy equivalence between compact PL manifolds (or smooth manifolds) cannot be homotopic to a PL (or smooth) homeomorphism if its torsion is nontrivial. (Later, the topological invariance of Whitehead torsion allowed one to eliminate the "PL" and "smooth" in all of that, extending these tools to, for example, topological manifolds without using triangulations.)
But the $h$-cobordism theorem says more: it applies Whitehead's invariant to manifolds in a different and deeper way.
Meanwhile on the $K_0$ side Wall introduced his invariant to detect whether there could be a finite complex in a given homotopy type. Note that where $K_0$ is concerned with existence of a finite representative for a homotopy type, $K_1$ is concerned with the (non-)uniqueness of the same.
Siebenmann in his thesis applied Wall's invariant to a manifold question in a way that corresponds very closely to the $h$-cobordism story: The question was, basically, when can a given noncompact manifold be the interior of a compact manifold-with-boundary? Note that there is a uniqueness question to go with this existence question: If two compact manifolds $M$ and $M'$ have isomorphic interiors then this leads to an $h$-cobordism between their boundaries, which will be a product cobordism if $M$ and $M'$ are really the same.
One can go on: The question of whether a given $h$-cobordism admits a product structure raises the related question of uniqueness of such a structure, which is really the question of whether a diffeomorphism from $M\times I$ to itself is isotopic to one of the form $f\times 1_I$. This is the beginning of pseodoisotopy theory, and yes $K_2$ comes into it.
But from here on, the higher Quillen $K$-theory of the group ring $\mathbb Z\pi_1(M)$ is not the best tool. Instead you need the Waldhausen $K$-theory of the space $M$, in which basically $\mathbb Z$ gets replaced by the sphere spectrum and $\pi_1(M)$ gets replaced by the loopspace $\Omega M$. It's a long story!
No, the answer is negative in general (if you require $M$ to be compact). $M$ comes with a map $M \to BG$ that is, by definition, $n+1$-connected (iso on $\pi_i$ for $i=0,...,n$, epi on $\pi_{n+1}$). You can turn it into a weak equivalence by attaching cells of dimension $\geq n+1$. From that you see, that there is a model for $BG$ having finite $n$-skeleton. This is a special property of a group that is called $F_n$ (for more information, see http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/). Finitely presented groups are $F_2$ and you find that a necessary condition on your $G$ is that it is of type $F_n$. The are concrete examples of groups that are $F_i$ but not $F_{i+1}$ for each $i$, which are discussed in same blog post (on page 423 in Hatcher's AT, you find the same examples in a slightly different context).
On the other hand, let $G$ be $F_n$ and let $K$ be the $n$-skeleton of $BG$; a finite complex. Then I claim there is a closed manifold $M$ with the desired properties. $M$ can be chosen of arbitrary dimension $d \geq 4,2n+1$ and to be stably parallelizable. Start with a sphere $S^d \to K$ and do surgery on $S^d$ to get rid of the homotopy groups in low dimensions. The precise formulation is for example Proposition 4 in Kreck's paper "Surgery and duality".
So we can say that a necessary and sufficient condition is that $G$ is of type $F_n$.
Caveat: I might have confused $n$ and $n+1$ at various places.
If you want to have $dim M \leq 2n$, you meet a new obstruction enforced by Poincare duality and things become really difficult.
Best Answer
Update:
My memory was quite blurry about this when I originally answered.
See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseignement Mathématique (2) 56 (2010), 143-171.
They prove that any finitely presented group is a subgroup of the fundamental group of the complement of a closed orientable surface in the $4$-sphere, which is much better than I reported.
Original answer:
You most likely would like a finitely presented group, but this might be of interest anyway:
Let $S$ be a recursively enumerable non-recursive subset of the natural numbers and consider the group
$\langle \ a,b,c,d \ | \ a^iba^{-i} = c^idc^{-i} \ \mathrm{for}\ i \in S \rangle$
This has unsolvable word problem. See page 110 of Chiswell's book "A course in formal languages, automata and groups" available on google books (I think it's also in Baumslag's "Topics in Combinatorial Group Theory" but all my books are in boxes at the moment.)
This should be the fundamental group of the complement of a noncompact surface in $\mathbb{R}^4$. You do this in the usual way by beginning with the trivial link on four components and then drawing the movie of the surface in $\mathbb{R}^4$, band summing at each stage to make the conjugates of $b$ and $d$ equal.
I think you end up with a knotted disjoint union of planes. I remember doing this at some point in graduate school when C. Gordon asked me if there were any compact surfaces in the $4$-sphere whose complements have groups with unsolvable word problem.