Using Stokes' theorem to define "area" enclosed by a curve

Yes, this is a lower bound on the area of a surface bounded by the curve. Parameterize the surface and apply Stokes' theorem. Your squared area is:

$$\sum_{1 \leq i < j \leq n} \left( \int_S dx_i dx_j \right)^2 = \int_S \int_S \sum_{1\leq i < j \leq n} dx_i dx_j dy_i dy_j$$

whereas the actual squared area is:

$$ \left( \int_S \sqrt{ \sum_{1 \leq i < j \leq n} \left( dx_i dx_j \right)^2} \right)^2= \int_S \int_S \sqrt{\sum_{1 \leq i < j \leq n} \left( dx_i dx_j \right)^2} \sqrt{\sum_{1 \leq i < j \leq n} \left( dy_i dy_j \right)^2}$$

which is at least as large by Cauchy-Schwartz. Equality occurs exactly when the area element of the surface is always pointed in the same direction in $\wedge^2 \mathbb R^n$ - that is, when the surface is flat.

This also shows that your definition is coordinate-invariant even before you average over $SO_n$.

The answer to my question in the edit is negative. Following the remark of Liviu Nicolaescu, one can always take the figure 8, embed it in the $x_1x_2-$plane and do a small perturbation such it doesn't intersect it self. Doing so you can get a "Stokes area" as small as you like, no matter how big your figure 8 is.

This is not the answer to your question, but a related result. If $\gamma(x)=(x(s),y(s))$ is a closed curve on the plane, then $$ \mathfrak{D}=\frac{1}{2}\int_0^1(\dot{y}(s)x(s)-\dot{x}(s)y(s))\, ds $$ is the oriented area. In particular if the curve has the shape of the digit $8$, the oriented area equals zero because the curve has opposite orientation in the upper and the lower part of $8$.

In the case of closed curves in $\mathbb{R}^{2n}$ with coordinates $x_1,y_1,\ldots,x_n,y_n$, one can define the symplectic area as the sum of oriented areas of projections onto the $x_iy_i$-planes, $i=1,2,\ldots,n$. More precisely, the symplecic area is $\mathfrak{D}=\mathfrak{D}_1+\ldots+\mathfrak{D}_n$, where $$ \mathfrak{D}_j=\frac{1}{2}\int_0^1(\dot{y}_j(s)x_j(s)-\dot{x}_j(s)y_j(s))\, ds $$ is the oriented area of the projection of the curve on the $x_iy_i$-plane.

Theorem (Isoperimetric inequality). If $\gamma=(x_1,\ldots,x_n,y_1,\ldots,y_n):[0,1]\to\mathbb{R}^{2n}$ is a closed rectifiable curve prametrized by constant speed, then the following isoperimetric inequality is true $$ L^2\geq 4\pi|\mathfrak{D}|, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*) $$ where $L$ is the length of $\gamma$ and $\mathfrak{D}=\mathfrak{D}_1+\ldots+\mathfrak{D}_n$ is defined above. Moreover the equality in (*) holds if and only if $\gamma$ is a circle of the following form: there are points $A,B,C,D\in\mathbb{R}^n$ such that form \begin{equation} \label{Eq1} \gamma(s)=(C+iD)+(1-e^{+2\pi is})(A+iB), \quad \text{when} \quad L^2=4\pi\mathfrak{D} \end{equation} and \begin{equation} \label{Eq2} \gamma(s)=(C+iD)+(1-e^{-2\pi is})(A+iB) \quad \text{when} \quad L^2=-4\pi\mathfrak{D}. \end{equation}

The proof follows from a simle and straightforward adaptation of a standard proof of the isoperimetric inequality due to Hurwitz, see:

P. Hajłasz, S. Zimmerman, Geodesics in the Heisenberg group. Anal. Geom. Metr. Spaces 3 (2015), 325–337.