Volume form induced by a metric I know that a Riemannian manifold of dimension $n$ is naturally endowed with a volume form induced by the Riemannian metric $\omega=\sqrt{|g|}dx_1\wedge\dots\wedge dx_n$.
Is the same thing true with only topological assumptions? 
With this I mean: does a metric induce something similar to a volume form in a metric space?

Let me first try to make sense of your question. First of all, the notion of a differential form is meaningless for general metric spaces $(X,d)$. However, one can still talk about Borel measures $\mu$ on the topological space $X$ (topologized using the metric $d$). The Borel condition still leaves too much freedom since we did not use the metric $d$ (only the topology). The most common condition these days is the one of a metric measure space which ties nicely $d$ and $\mu$ and allows one to do quite a bit of analysis on $(X,d,\mu)$ similarly to the analysis on the Euclidean $n$-space $E^n$. (The literature on this subject is quite substantial, just google "metric measure space".) 
Definition. A triple  $(X,d,\mu)$ (where $d$ is a metric on $X$ and $\mu$ is a Borel measure on $X$) is called a metric measure space if the measure $\mu$ is doubling with respect to $d$, i.e. there exists a constant $D<\infty$ such that for every $a\in X, r>0$, we have
$$
\mu(B(a, 2r))\le D \mu(B(a,r)),
$$
where $B(a,R)$ denotes the closed ball of radius $R$ centered at $a$. 
Example. Every closed subset of $E^n$ (equipped with the restriction of the Euclidean metric) is a complete doubling metric space. 
The basic existence result for doubling measures $\mu$, proven in 
"Every complete doubling metric space
carries a doubling measure", by J. Luukainen,  E. Saksman, Proc. Amer. Math. Soc. 126 (1998) p. 531–534, is the following:
Theorem. A complete metric space $(X,d)$ carries a doubling measure $\mu$ if and only if the metric $d$ is doubling, i.e. there exists a constant $C$ such that every ball of radius $2r$ in $X$ is covered by at most $C$ balls of radius $r$.  
The proof of this theorem is not long but by no means trivial. 
