Using Stokes' theorem to define "area" enclosed by a curve I am trying to figure out what the next calculation of the "area" (or "volume" in higher dimensional analogues) using Stokes' theorem really means. Here is my thought process: 
$2$-dimensional case: given a closed simple piece-wise smooth curve $C$ in $\mathbb{R}^2$ you can find out the area enclosed by $C$ using Green's theorem by choosing an orientation on $C$ and calculating 
$$\text{Area} = \left\vert \dfrac{1}{2} \oint_C (x dy - y dx) \right\vert .$$ 
$2$-dimensional case in $\mathbb{R}^n$: When you have a piece-wise smooth non self intersecting map $\phi : S^1 \rightarrow \mathbb{R}^n$ such that $\phi (S^1)$ is contained in a $2$-dimensional plane of $\mathbb{R}^n$, you can again find the area enclosed by $\phi (S^1)$ by choosing an orthonormal coordinate system $(x_1,...,x_n)$, choosing an orientation on $S^1$ and calculating using Stokes' theorem
$$\text{Area}^2 = \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 .$$ 
Now my question is what is the meaning of calculating the above quantity when $\phi (S^1)$ is not contained in a $2$-dimensional plane. I want to use the same computation as above, but since I want the number I get to be independent on the coordinate system, I'll have to average with respect to all coordinate systems. To be explicit: let  $\phi : S^1 \rightarrow \mathbb{R}^n$ be a piece-wise smooth non self intersecting map and fix an orientation on $S^1$. Let $SO_n (\mathbb{R})$ be the special orthogonal group and let $\mu$ be the normalized Haar measure on $SO_n (\mathbb{R})$ (i.e., $\mu (SO_n (\mathbb{R}) ) =1$). Define the "area" (or the "Stokes area" in lack of a better name) bounded by $\phi (S_1)$ as
 $$\text{"Stokes area"}^2 = \int_{SO_n (\mathbb{R})} \left( \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\gamma.\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2  \right) d \mu (\gamma) .$$
It is not hard to give analogues in any dimension to get "Stokes volume" (note that the analogue in dimension $1$, i.e., for $S^0$, comes out just the usual Euclidean distance between points).
My questions are: 


*

*Is this quantity well-known / studied ? If so, I would very much appreciate a reference. 

*What does this "Stokes area" represent geometrically (I have no intuition about it)? 

*Is there a connection (say by some inequality) to the area of a minimal surface enclosed in $\phi (S^1)$? 
Edit: by the answer of Will Sawin below, it is clear that we can consider only 
$$\text{"Stokes area"}^2 =  \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\gamma.\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 .$$
and that is quantity is always less or equal than the area of a surface enclosed by the curve. The question remains is there is a connection to the infimum of the areas of all the surfaces enclosed by the curve. In the case the curve is contained in a plane, the formula computes this minimum. But what about the general case: is there a constant (maybe depending on the dimension of $\mathbb{R}^n$) $c(n)$ such that 
$$ \text{"Stokes area"} \geq c(n) (\text{infimum of the areas of surfaces enclosed by the curve}) ?$$
 A: 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.  
A: 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.
A: 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 $SO_n$.
