Using Cholesky decomposition, $\Sigma = L L^T$. Now define $y$ such that $L y = x - \mu$. Then:
$$ \begin{eqnarray} (x - \mu)^T \Sigma^{-1} (x - \mu) &=& (L y)^T (L L^T)^{-1} (L y) \\ &=& L^T y^T (L^T)^{-1} L^{-1} L y \\ &=& y^T y \\ &=& \sum y_i^2 \\ &=& \chi^2 \end{eqnarray} $$Because, as we previously showed, each component of $y$ is an independent t-value.
Just for fun, let's verify the result with some code.
import matplotlib.pyplot as plt
import numpy as np
import scipy.stats
np.random.seed(42)
Draw MVN samples.
dim = 3
mu = np.random.random(dim)
covar = np.array([[ 2, -1, 0],
[-1, 2, -1],
[ 0, -1, 2]])
samples = np.random.multivariate_normal(mu, covar, 250000)
Compute $(x - \mu)^T \Sigma^{-1} (x - \mu)$ for each sample.
covar_inv = np.linalg.inv(covar)
vals = np.array([np.dot(np.dot((x - mu).T, covar_inv), x - mu) for x in samples])
Plot the results.
plt.hist(vals, bins=100, range=(0, 14), normed=True, label='$(x - \mu)^T \Sigma^{-1} (x - \mu)$')
x = np.linspace(0, 14, num=1000)
y = scipy.stats.chi2.pdf(x, dim)
plt.plot(x, y, label='Chisquare({0})'.format(dim), color='red', linewidth=3)
plt.legend()
<matplotlib.legend.Legend at 0xb2e30ec>