Think back to what the Riemannian volume form is. It's a form that, when fed an oriented orthonormal frame, spits out 1. (Of course every manifold here is oriented.)
What are the oriented orthonormal frames on $\Sigma \subset M$, where $\Sigma$ is a (Riemannian) submanifold? Because it's a Riemannian submanifold, if $(x_1, \dots, x_{d-1})$ is an oriented orthonormal frame of $T_p \Sigma$, it's also orthonormal in $T_p M$ - we're just missing one vector. By definition, $n$ is normal to all of these, and (modulo orientation conventions), $(n, x_1, \dots, x_{d-1})$ is an oriented orthonormal frame of $T_p M$. Therefore $$i_nV_\Sigma(x_1, \dots, x_{d-1}) = V_M(n, x_1, \dots, x_{d-1}) = 1,$$ so that $i_nV_\Sigma$ does exactly as a volume form should.
You cannot refer to the induced volume form without at least having chosen a nonvanishing normal field. The Riemannian metric gives you a Riemannian structure on the normal bundle of $\Sigma$, which is what you need to define the induced volume form on $\Sigma$.
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.
Best Answer
I think that in general the best approach is the following: For all this discussion, we start with a Riemannian metric $ds^2$ on $M$, and we look at the induced Riemannian metric $i^*ds^2$ on $N$. We write $$i^*ds^2 = \sum_{j=1}^n \omega^j\otimes\omega^j$$ for a suitable collection of $1$-forms $\omega^j$. Then the induced volume ("area") form on $N$ will be $\omega^1\wedge\dots\wedge\omega^n$.
For example, consider $S^2\hookrightarrow \mathbb R^3$. Considering spherical coordinates, $i(\phi,\theta) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$, we have \begin{align*} i^*ds^2_{\mathbb R^3} &= i^*\big(dx^1\otimes dx^1+ dx^2\otimes dx^2+dx^3\otimes dx^3\big) \\ &= d\phi\otimes d\phi + \sin^2\phi\, d\theta\otimes d\theta \\ &= \omega^1\otimes\omega^1 + \omega^2\otimes\omega^2\,, \end{align*} where $\omega^1 = d\phi$ and $\omega^2 = \sin\phi\,d\theta$. [We order these to give the orientation we want on the submanifold.] Then our area form on $S^2$ is $$\omega^1\wedge\omega^2 = \sin\phi\,d\phi\wedge d\theta\,.$$