Short answer: The measure is the higher dimensional analogue of arclength and surface area. It's not Lebesgue measure from the ambient Euclidean space directly. But $\mathbb{R}^N$ is an inner product space, and that is used to determine lengths of vectors, volume of boxes, and Lebesgue measure on $\mathbb{R}^N$. Any submanifold of $\mathbb{R}^N$ inherits that inner product in the form of a Riemannian metric, which determines lengths of tangent vectors, volume of “infinitesimal” (i.e. tangent) boxes, and by integrating, a measure.
At more length:
No, the measure on the boundary $\partial\Omega$ is not Lebesgue measure on $\mathbb{R}^{N-1}$. If $\Omega$ is an $N$-dimensional domain in $\mathbb{R}^N$, then its boundary $\partial\Omega$ is an $(N-1)$-dimensional submanifold of $\mathbb{R}^N.$ It is not a subset of $\mathbb{R}^{N-1}$, so its measure cannot be computed using Lebesgue measure on $\mathbb{R}^{N-1}$.
In general, a $k$-dimensional submanifold of $\mathbb{R}^N$ need not be naturally a subset of any $\mathbb{R}^k$, so we can't use Lebesgue measure on $\mathbb{R}^k$ to measure it.
However, a $k$-manifold is locally homeomorphic to $\mathbb{R}^k$, which means that each local patch has a $\sigma$-algebra isomorphic to that of $\mathbb{R}^k$. Does this give us a measure we can use for the $k$-dim submanifolds of $\mathbb{R}^N$? No, a submanifold may be topologically nontrivial, like a 2-sphere, so that it requires more than one coordinate patch to cover it. A set may be in more than coordinate patch, and what measure you assign it depends on which coordinates you use. So that's not well-defined.
But to state the problem is to solve it. When you change coordinate patches, the measure as defined in $k$-dimensional Euclidean space transformed by the Jacobian of the transition map. If you can account for that transformation, you will get a well-defined measure.
Here it is in more detail, but what follows is a straightforward generalization of the notions of arc-length and surface area from multivariable calculus, using a bit of the language of linear algebra and Riemannian geometry.
First, the linear algebra. Given an inner product space $V$, we assign a vector $v$ a length $\lVert v\rVert = \sqrt{\langle v,v\rangle}.$ We represent parallelograms spanned by two vectors $u,v$ by their wedge product $u\wedge v$ in the exterior algebra on $V$. We can extend the inner product to these paralellograms (also known as bivectors or 2-planes):
$$\langle u\wedge v, x\wedge y\rangle=\langle u,x\rangle\langle v,y\rangle-\langle u,y\rangle\langle v,x\rangle.$$
Then the area of a parallelogram is the magnitude of the corresponding bivector:
$$\text{area}(u,v)=\lVert u\wedge v\rVert = \sqrt{\langle u\wedge v,u\wedge v\rangle}.$$
Similarly, for any $k<N$, we consider $k$-dimensional paralellepipeds aka $k$-planes, which we represent as $k$-vectors, which are wedges of $k$ different vectors. We have the general formula for the inner product of $k$-vectors as a Gram determinant:
$$\langle u_1\wedge\dotsc\wedge u_k, v_1\wedge\dotsc\wedge v_k\rangle=\det[\langle u_i,v_j\rangle]$$
and
$$\text{vol}_k(u_1,\dotsc,u_k)=\lVert u\wedge \dotsb\wedge u_k\rVert = \sqrt{\langle u\wedge \dotsb\wedge u_k,u\wedge \dotsb\wedge u_k\rangle}.$$
Now that we know how to compute the volumes of boxes, we slice our $k$-dimensional manifold into infinitesimal pieces, compute the size of each piece as the volume of the $k$-plane spanned by the tangent vectors. We sum the volumes, Riemann integral-style, to get the total measure. In other words, if our $k$-manifold $M$ is parametrized as $\mathbf{x}(u_1,u_2,\dotsc,u_k)$, the tangent space is spanned by $\partial\mathbf{x}/\partial u_1,\dotsc, \partial\mathbf{x}/\partial u_k$. The infinitesimal tangent box has volume $$\lVert\partial\mathbf{x}/\partial u_1\wedge\dotsb\wedge\partial\mathbf{x}/\partial u_k\rVert$$ and the measure of the submanifold is given by
$$\text{vol}_k(M) = \int_M\lVert\partial\mathbf{x}/\partial u_1\wedge\dotsb\wedge\partial\mathbf{x}/\partial u_k\rVert\,du_1\,\dotsc\,du_k.$$
And that's the measure. Put $k=N-1$ to compute the measure of the $(N-1)$-dimensional submanifold $\partial\Omega\subseteq\mathbb{R}^N$.
To clarify some terminology, any function which assigns numbers to $k$-planes (and subject to certain transformation rules to ensure it's well-defined which make it a density ) can be integrated, Riemann integral style. If these functions are also linear, they are called differential forms, but the ones we're considering are not linear, so they're only densities.
The integral of the density gives the measure.
Best Answer
I'm not going to prove the divergence theorem here, because you can find proofs of this in books which cover integration on manifolds. Particularly, Amann and Escher's Analysis III covers all the necessary measure-theory and Lebesgue-integration background. In the end they also have a chapter dedicated to integration on manifolds (where Stokes theorem is proved in great generality, allowing for slightly rougher boundaries than what is usually proven, for example in Spivak's Calculus on manifolds. Of course, more analysis-related technicalities are present to make all the approximations work out).
What I will mention is the very basics of measures on manifolds. Throughout, $M$ shall denote a $C^r$ ($r\geq 1$), $m$-dimensional manifold, which we assume to be second-countable, which means we have a countable atlas (this is automatically true for submanifolds of $\Bbb{R}^n$, like $\partial \Omega$ in your case).
1. What is 'the' $\sigma$-algebra?
Well, $M$ is a $C^r$ manifold, so it is in particular a topological space, hence we can equip it with the Borel $\sigma$-algebra.
Alternatively, we can also equip it with the Lebesgue $\sigma$-algebra $\mathcal{L}_M$. The definition is that a subset $A\subset M$ belongs to $\mathcal{L}_M$ if for every chart $(U,\alpha)$ of $M$, we have that $\alpha[A\cap U]$ is a Lebesgue-measurable subset of $\Bbb{R}^m$. You can easily verify that the collection $\mathcal{L}_M$ is indeed a $\sigma$-algebra, which contains the Borel $\sigma$-algebra.
2. Volume Measures on Pseudo-Riemannian Manifolds.
Let $(M,g)$ be a pseudo-Riemannian manifold (of class $C^r$, $r\geq 1$, of dimension $m$, and again, second-countable). There is a very important measure $\lambda_g$ which can be defined on the Lebesgue-$\sigma$-algebra of $M$, and it is constructed from the metric tensor field $g$. This $\lambda_g$ captures our intuitive understanding of volumes of manifolds. The precise theorem is as follows:
To prove this, note that for each chart $(U,\alpha)$ belonging to the maximal atlas, we can define $\lambda_{g,(U,\alpha)}(A):= \int_{\alpha[A]}\sqrt{|\det [g_{(\alpha)}]|}\,d\lambda_m$. This defines a positive measure on $\mathcal{L}_M|_{U}$ (by monotone convergence) which is complete (since the standard Lebesgue measure $\lambda_m$ is, and since the integrand is strictly positive) and $\sigma$-finite (since the standard Lebesgue measure $\lambda_m$ is). Also, if we consider two different charts $(U,\alpha)$ and $(V,\beta)$, then in the overlap, these measures agree by a simple application of the change-of-variables theorem. Therefore, applying theorems 1 and 2 stated below gives us the desired $\lambda_g$.
Btw this result is also given in the book, though there it is proven using a partition of unity argument... but I prefer this more elementary approach, hence I presented this. In fact, we can generalize this and get a measure $\lambda_{\rho}$ from any non-negative, measurable, scalar density $\rho$. As a brief summary: a scalar density is essentially an object $\rho$ which to each chart $(U,\alpha)$ gives a function $\rho_{\alpha}:\alpha[U]\to\Bbb{R}$, such that for two different charts $\rho_{\beta}$ and $\rho_{\alpha}$ differ by a Jacobian determinant. This determinant is exactly the thing needed to cancel the Jacobian from the change-of-variables theorem, thereby giving us a whole bunch of compatible measures $\lambda_{\rho,(U,\alpha)}$ to which we can apply theorem 1 below (and by imposing more conditions on the scalar density, we can apply theorem 2).
3. Usual Applications.
Usually, what happens is that we're given a (say Riemannian) manifold $(M,g)$, along with a submanifold $S$ of $M$. Then, the standard way of inheriting a metric tensor on $S$ is by taking the pullback $\iota^*g$, where $\iota:S\to M$ is the canonical injection. Then, by the theorem above, we have a corresponding measures $\lambda_g$ on $M$ and $\lambda_{\iota^*g}$ on $S$. In the case when $S$ has codimension $1$ in $M$, we refer to $\lambda_{\iota^*g}$ as the induced surface measure on $S$, and very often it is denoted as $\sigma$. So, volume integrals over $M$ are written $\int_M f\,d\lambda_g$ or just $\int_Mf \,dV$, while integrals over the submanifold $S$ are written $\int_Sf\,d\lambda_{\iota^*g}$ or just $\int_Sf\,d\sigma$, or maybe even $dS$ is written.
As a hyper special case, consider $M=\Bbb{R}^n$, with the usual Riemannian metric $g=\sum_{i=1}^ndx^i\otimes dx^i$, where $(x^1,\dots, x^n)$ is the usual coordinate system (i.e corresponding to the identity chart, or simply what we call cartesian coordinates). Then, we can consider any submanifold $S$ of $\Bbb{R}^n$, for example a sphere of certain radius, or for example in $\Bbb{R}^3$, we can consider a certain sheet of the hyperboloid $x^2+y^2-z^2=1$ etc etc. I your case, we can consider $M=\Omega$ to be an open set with nice boundary in $\Bbb{R}^m$, and $S=\partial\Omega$. Then, all the above machinery applies here.
If you know about differential forms, then you may know that on any oriented Riemannian manifold, one can define a volume form (i.e a top-degree, nowhere vanishing differential form) $\mu_g$, which has the property that on every positively oriented, $g$-orthonormal basis of each tangent space, $\mu_g$ evaluates to $1$. In this case, the integration of a function $f:M\to\Bbb{R}$ with respect to the measure $\lambda_g$ is the same thing as the integration of the top degree differential form $f\,\mu_g$ (so all the notions are consistent; though using $\lambda_g$ is more general since it doesn't require orientability).
Measure-Theoretic Facts
Here are the necessary measure-theoretic facts used above.
To prove this we essentially follow our noses.
This completes the proof of the theorem.
Now, we can ask what properties the measure $\mu$ inherits from the $\{\mu_i\}_{i\in I}$. Here’s an indication of results in this direction
Note that we could investigate many other properties as well, such as how regularity (i.e approximability) properties of each $\mu_i$ implies that for $\mu$. Another thing to investigate would be what happens if we have another collection of measures $\{\nu_i\}_{i\in I}$ (same indexing set) which also satisfies theorem 1, and such that for each $i\in I$ (or simply $i\in I_0$) we have that $\nu_i$ and $\mu_i$ are $\sigma$-finite and $\nu_i\ll \mu_i$. Then, we could investigate whether or not the patched measures are also absolutely continuous $\nu\ll\mu$, and in this case we can ask how the Radon-Nikodym derivatives are related. Another thing to ask would be to consider a completely different family of $\sigma$-finite measures $\{(Y_j,\nu_j)\}_{j\in J}$, and ask how the product measure $\mu\times \nu$ is related to each of the $\mu_i\times \nu_j$ etc. The point is that once you have the ‘big theorems’ of Fubini and Radon-Nikodym, you can easily patch these results together and investigate what happens.