Consider these three polygons:

Object #1 is clearly symmetrical. Object #2 is not, but in some sense it seems "close" to being so (in that it is not very different from object #1). Object #3 appears to be very far from being symmetrical.
Is there a standard way of quantifying these observations, i.e., a measure for "closeness to symmetry"?
Let's pretend we have defined rigorously what we mean by a symmetric shape. Let's also pretend we have some sensible metric metric on the set of shapes in $\mathbb R^2$. Then we can measure how symmetric the shape $A$ is using the non-negative number. . .
$|A|= \inf \{d(A,B): B$ is symmetric$\}$
Symmetric shapes will always satisfy $|A|=0$ because $d(A,A)=0$. Whether there are non symmetric shapes will depend on the measure and the exact notion of symmetry.
So what kind of metrics could we use? Well there's always the Hausdorff metric on $\mathbb R^2$: For any two shapes (subsets) $A$ and $B$ we define the distance. . .
$d(A,B) = \inf \{\varepsilon : $every point of $A$ is within distance $\varepsilon$ of some point of $B$ and vice-versa$\}$
This is a good metric because according to it a 'shape' can be any closed subset of the plane. However the Hausdorff metric might give $(1)$ and $(3)$ as being very far apart because the tip of $(3)$ is very far from the edge of the square would be.
Another metric for shapes is by taking the symmetric difference. . .
$A \Delta B = \{x \in \mathbb R^2 : x$ is an element of exactly one of $A$ or $B$ $\}$
and calculating its area:
$d(A,B) =$ The area of $A \Delta B$
This metric is more likely to give $d((1),(3))< d((1),(2))$. However it only allows for a narrower definition of what a 'shape' is. For example a disc will be distance $0$ from a 'lollipop' shape because the symmetric difference is just the stick, which has measure zero. So we might restrict to, for example, finite unions of polygons. Maybe that's enough for what you want.
Edit: For either of the two metrics you probably want to be allowed move the shapes around before comparing them. In that case you actually want to consider the distance. . .
$D(A,B) = \inf \{d(A,B+x): x \in \mathbb R^2\}$
where $B+x$ just means $\{b + x : b \in B\}$