There is a simple proof of Gauss-Green theorem if one begins with the assumption of Divergence theorem, which is familiar from vector calculus,
\begin{equation}
\int_{U}\mathrm{div}\,\mathbf{w}\,dx =
\int_{\partial U} \mathbf{w}\cdot\mathbf{\nu}\,dS,
\end{equation}
where $\mathbf{w}$ is any $C^\infty$ vector field on $U\in\Bbb{R}^n$ and $\mathbf{\nu}$
is the outward normal on $\partial U$.
Now, given the scalar function $u$ on the open set $U$, we can construct the vector field
\begin{equation}
\mathbf{w}=(0,\ldots,0,u,0,\ldots,0),
\end{equation}
where $u$ is the $i$th component. Then, following the Divergence theorem,
we have
\begin{equation}
\int_U \mathrm{div}\,\mathbf{w}\,dx=\int_U u_{x_i}\,dx
=\int_{\partial U}\mathbf{w}\cdot\mathbf{\nu}\,dS
=\int_{\partial U}u\nu^i\,dS.
\end{equation}
In Evans' book (Page 712), the Gauss-Green theorem is stated without proof and the Divergence theorem is shown as a consequence of it. This may be opposite to what most people are familiar with.
Notice $\partial \Omega$ is a Riemannian hypersurface of $\mathbb R^n$. The measure on $\partial \Omega$ is the one given by the Riemannian metric.
Let me define the measure more explicitly.
Since $\Omega$ is open (and therefore orientable), $\partial \Omega$ is orientable. Knowing $\partial \Omega$ is an orientable hypersurface we have a global normal vector field $n$. We can suppose $|n(p)|=1$ for all $p\in \partial \Omega$.
For each $p$ define $$\eta(v_1,\ldots,v_{n-1}) = \det(n(p),v_1,\ldots,v_{n-1})$$
where $v_1,\ldots,v_{n-1} \in T_p \partial \Omega$. This $\eta$ is the Riemannian volume form of $\partial \Omega$.
The measure you want is the one provided by this volume form. More precisely you can obtain the measure using the Riesz–Markov–Kakutani representation theorem on the functional
$C(\partial \Omega) \to \mathbb R$
given by
$$f \mapsto \int_{\partial \Omega}f \eta.$$
Best Answer
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.