Is there a parameter which describes the degree of isotropy/anisotropy of a single source? Our task is to analyse how isotropic a random collection of vectors is. All of them start at the origin and have the same length. Is there a parameter which describes the isotropy for this case? How do I calculate it? 
 A: Probably an histogram (with the appropriate distribution depending on the solid angle) as suggested by @Gabriel Golfetti  would be enough.
As an additional approach, I can suggest using Principal Component Analysis which basically tells you if there is a "direction" in space (2D, 3D, 4D...) which is preferred. 
It requires some coding and mathematical capabilities but is a very general concept. I am writing this for the sake of completeness.
You basically take your vectors, assume the components of the vector are position in space (i.e. if $v=(3, 4, 2)$ then $x=3$, $y=4$, $z=2$) so you have a collection of points in space. Compute the covariance matrix of those points, find its eigenvectors and the relative eigenvalues. Compare the eigenvalues. If they are all the same, your collection is isotropic. Otherwise, it is mostly "spread" in the direction with the bigger eigenvalue (i.e. along the eigenvector with the bigger eigenvalue) or rather, the bigger the difference between eigenvalues, the bigger the spreading of the value over the direction with the bigger one.
Taking inspiration from this weird wiki page I think the parameter you need in 3D is, if you have eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$:

which is 0 for isotropic distributions and 1 if the distribution is all along a single direction.
I think the N-Dimensional exchange (when you have N eigenvalues) would be:
$$A { \sqrt{ \Sigma_{(i, j), i!=j} (\lambda_i-\lambda_j)^2}  \over \sqrt{\Sigma_i^N \lambda_i^2} }$$
with $A$ some constant assuring that the maximum is 1 I think, and it should be $\sqrt{1/m}$ where $m$ is the number of times each eigenvalue appears in the sum on the top.
Notice that this procedure is fooled if all the vectors are only aligned along orthogonal  directions (e.g. of $n$ vectors, $n/3$ along $x$, $n/3$ along $y$ and $n/3$ along z) which gives FA=0. This distribution would indeed in some sense be isotropic (no preferred direction) but not uniform in space (for that, chech the histograms).
A: I think the problem you have is 3d. I'm going to outline a method in 2d however and mention how to generalize it. 
For convenience, say the vectors all have length 1. Write their components down in polar coordinates. You will have a set of vectors
$$\mathbf v=\cos\theta\hat x+\sin\theta\hat y$$
Notice how $\theta\in[0,2\pi)$ uniquely specifies the direction of the vector. 
Now you should make a histogram of all these $\theta$, and see how uniform the distribution is. If it varies too much from an uniform distribution, the source is anisotropic. 
To make this 3d, instead you would have to plot this for general solid angles, and it's a bit more complicated. Directions can be parametrized in spherical coordinates, but your histogram is going to need some different weights depending on the polar angle as to take into account the different sizes of the circles as you vary that angle. 
If you want a single parameter to measure that, you could look into some statistics. Take for example https://stats.stackexchange.com/questions/25827/how-does-one-measure-the-non-uniformity-of-a-distribution and answers therein. 
A: One can calculate the standard deviation of multiple vectors using circular statistics.
Here, you're essentially attaching the vectors tip-to-tail and asking whether the result—i.e., the resultant vector—is long or short. If short, then the variance is high, and the vectors are relatively anisotropic. If long, then the variance is low, and the vectors are relatively isotropic. 
When the vector direction is uniform (i.e., the vectors are perfectly isotropic), then the length of the resultant vector is maximized (in your case, it would simply be $NL$, where $N$ is the number of vectors of length $L$).
It seems like one could extend this strategy to the 3-D case.
