Let $X\sim N(\mu,\Sigma)$ be a random variable in $\mathbb{R}^d$ with multivariate normal distribution. Let $\hat\Sigma$ be the maximum likelihood estimator for $\Sigma$,
\begin{align} \hat\Sigma=\frac1n\sum_{i=1}^n(\vec{x}_i-\bar x)(\vec{x}_i-\bar x)^T, \end{align}
where $\vec{x}_i\in\mathbb{R}^d$ is the $i$th sample vector and $\bar x\in\mathbb{R}^d$ is the sample mean of $X$.
What is a 95% confidence interval for $\det(\Sigma)$? That is, what is a confidence interval around $\det(\hat\Sigma)$ that will contain $\det(\Sigma)$ 95% of the time a sample is taken?
I was led to this question because I am trying to find confidence intervals for estimates of entropy and mutual information for multivariate normally distributed variables.
Edit: According to the article "The distribution of the determinant of a complex Wishart distributed matrix" by N.R. Goodman (see note at the end about the distibution of the determinant of a (real) Wishart distributed matrix, which is attributed to Wilks (1934)), the ratio $n^d(\det\hat\Sigma/\det\Sigma)$ is distributed as the product of $d$ independent $\chi^2$ random variables with $n, n-1, \cdots, n-d+1$ degress of freedom: \begin{align} \frac{\det\hat\Sigma}{\det\Sigma}\sim \prod_{k=0}^{d-1} \frac{Y_k}{n},\text{ where } Y_k\sim\chi^2_{(n-k)}. \end{align}
I checked this empirically with n=100, d=20, and it worked. The image at the bottom shows that the distribution of the variables on the left and right overlap almost identically. I simulated the variable on the left by choosing a fixed positive definite 20x20 covariance matrix $\Sigma$, and creating n=100 samples to find $\hat\Sigma$, and repeated $10^5$ times. I simulated the distribution on the right by sampling the appropriate $\chi^2$ distributions, dividing by n and forming the product ($10^6$ times because it was faster). In this case (n=100, d=20) the confidence interval for $\det\hat\Sigma/\det\Sigma$ is approximately $[0.0344,0.3120]$ (which is interesting since it doesn't include 1).
This means that if $l(n,d)$ is the .025-quantile, and $u(n,d)$ is the .975-quantile for $\prod_{k=0}^{d-1}\frac{Y_k}{n}$, then \begin{align} l(n,d) &\le \frac{\det\hat\Sigma}{\det\Sigma} \le u(n,d) \\ \frac{\det\hat\Sigma}{u(n,d)} &\le \det\Sigma \le \frac{\det\hat\Sigma}{l(n,d)} \end{align} 95% of the time.
So what I am hoping for is expressions for $l(n,d)$ and $u(n,d)$. Approximations in terms of n and d would be fine. (Let's say something like second order in $1/n$ and $1/d$ or any good approximation method that converges a lot faster than sampling the $\chi^2$ distributions).
Thank you! Any help would be greatly appreciated.