= Solution
The pivot $Z=\sqrt n\,\Sigma^{-1/2}(\overline X-\theta)$ has independent standard <normal coordinates>, so $Z^TZ$ has the <chi-squared distribution> with $d$ <degrees of freedom>. Let $q_{d,1-\alpha}$ be its $(1-\alpha)$ quantile. The <confidence ellipsoid>
$$
\boxed{C_n^\alpha=\left\{v\in\mathbb R^d:n(\overline X-v)^T\Sigma^{-1}(\overline X-v)\leq q_{d,1-\alpha}\right\}}
$$
therefore has exactly $1-\alpha$ coverage for every $\theta$. No large-sample approximation is required when the covariance is known.
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