From this paper by Cai, Fan, Jiang: you can calculate angles $\theta_{ij}$ between pairs of vectors $\vec{v}_i$, $\vec{v}_j$ from your collection, $1 \leq i < j \leq N$ (where $N$ is the total number of vectors). If the vectors have a uniform distribution on the $(n-1)$-sphere, then these angles should be distributed according to
$$
h(\theta) = \frac{1}{\sqrt{\pi}} \frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \cdot \left(\sin \theta\right)^{n-2}, \quad \theta\in[0,\pi],
$$
where $\Gamma(n)$ is the gamma function. This is a necessary condition for the uniform distribution on a sphere. And if I understand correctly, it is also sufficient (correct me if I am wrong - I am not an expert on this topic).
Here is a code in python that illustrates this.
import numpy as np
from numpy import sqrt, sin, arccos, pi
from numpy.linalg import norm
from scipy.special import gamma
import matplotlib.pyplot as plt
n = 15 # space dimension
N = 10000 # number of vectors
# Generate vectors (columns of v) uniformly distributed on the (n-1)-sphere
v = np.random.normal(0, 1, (n, N))
v_norms = norm(v, axis=0)
v = v / v_norms
# Calculate the angles between pairs of vectors
thetas = v.T @ v
thetas = thetas[np.triu_indices_from(thetas, 1)]
thetas = arccos(thetas)
# Plotting
fig, ax = plt.subplots(figsize=(14, 8), tight_layout=True)
ax.set_ylabel('pdf', fontsize=20)
ax.set_xlabel(r'$\theta_{ij}$', fontsize=20)
ax.hist(thetas, bins='auto', density=True, label='observed distribution')
x = np.linspace(0, pi, num=1000)
ax.plot(x, 1/sqrt(pi) * gamma(n/2)/gamma((n-1)/2) * (sin(x))**(n-2), color='red', linestyle='--', label=r'$\frac{1}{\sqrt{\pi}} \frac{\Gamma(n/2)}{\Gamma\left(\frac{n-1}{2}\right)} \left(\sin(\theta)\right)^{n-2}$')
plt.legend(loc='upper right', prop={'size': 18}, markerscale=4)
ax.set_xlim(0, pi)
plt.show()

P.S. Book "Symmetric Multivariate and Related Distributions" by Fang, Kotz, Ng (1990) is another relevant reading on this topic.