Let $M \subset \mathbb{E}^n$ be a submanifold, $\dim M = m$. Then $M$ inherits Riemannian structure from the Euclidean space. Consider the $\sigma$-algebra induced by the Lebesgue measurable sets and the parametrizations.
The surface measure on $M$ is defined as:
$$\sigma(A) = \int_{r^{-1}(A)} \sqrt{\det[g_{ij}]} d\lambda_m ~~~~~~(*) $$
where $g_{ij} = \langle\frac{\partial r}{\partial x_i}\mid\frac{\partial r}{\partial x_j} \rangle$, i.e. $[g_{ij}] = (Dr)^T (Dr)$ is the Gram matrix of $Dr$ or, in other words, the determinant of the inherited Riemann metric tensor. For simplicity, we assume here that the set $A$ lies within the image of a single local parametrization $r$.
This definition was motivated in a rather sloppy way:
Take a parametrization $r$ around $p \in M$ such that $r(0) = p$. Take a small hypercube $K \subset \mathbb{R}^m$ having $0$ as one of its vertices. Then the image $r(K)$ will approximately be a parallelogram contained in $TM_p$, with its sides parallel to the vectors $\frac{\partial r}{\partial x_i} \in TM_p$, whose volume is $\sqrt{\det g_{ij}} \lambda_m(K)$. Therefore, we define the measure as above.
I'm not convinced by an argument like this. It's completely unclear that the error of such approximation will ever converge to 0. Keep in mind that naive triangulation doesn't work - recall the Schwarz lantern. Moreover, the approximation error is not quantified at all.
I do agree that we only need to argue that the equality holds on images of hypercubes, as they form a generating set for the $\sigma$-algebra, so by the $\pi-\lambda$ lemma we'll get uniqueness of the measure satisfying $(*)$ for images of hypercubes.
Are there any different ways in which the surface measure is unique? Are there any limit constructions which justify the name surface area for a 2-dimensional manifold in $\mathbb{E}^3$? Can we justify the geometric meaning of the surface measure in a more rigorous way?
It is sad that such an important question has been left unanswered! So, let me attempt. People already mentioned change of variables formula, Hausdorff measure $H^n$, but here is the connection. First, I must say that I presume that we agree that $H^n$ deserves to be what "surface area" should be. So, my goal is to show that the particular choice/definition of $\sigma$ works because $\sigma = H^n$. The reason for the particular definition of Jacobian comes from linear algebra and looing into how linear maps translate $m$-volume into a bigger Euclidean space.
Lemma: Given any $\lambda>1$, $M$ can be covered by (countably many) local parameterizations $f\colon \mathbb{R}^m \supset U \to M$ such that
We may write $A$ as union of $A_i$ each contained in image of one of these local parameterizations. For each $A_i$, from (1), $$ H^n(A_i) \leq \lambda^n H^n(f^{-1}(A_i)), $$ But on $\mathbb{R}^n$, $H^n=\mathcal{L}^n$, so, $$ H^n(f^{-1}(A_i)) = \mathcal{L}^n(f^{-1}(A_i)) = \int_{f^{-1}(A_i)} 1 dx \leq \lambda \int_{f^{-1}(A_i)} J(f,x) dx = \lambda \sigma(A_i). $$ Here we used the fact that $\sigma$ has the same expression in any parameteriation, which is true because of the change of variables formula (for $C^1$ diffeomorphisms between subsets of Euclidean spaces).
So, $$ H^n(A_i) \leq \lambda^2 \sigma(A_i) $$ Similar argument gives $$ H^n(A_i) \geq \lambda^{-2} \sigma(A_i). $$ By additivity, we get same estimates for $A$. Since $\lambda>1$ was arbitrary, we get $$ H^n(A) = \sigma(A). $$