I am not sure the following helps.

If $f$ is a prob.distribution on $\mathbb{R}^n$ having the same covariance matrix as $N(0, I_n)$, then \begin{eqnarray*}D(f\|N(0, I_n))&=& H(N(0, I_n))-H(f)\\\ &=& \frac{1}{2}\log((2\pi e)^n)-H(f)\end{eqnarray*} where $D(\cdot \| \cdot)$ denotes the KL divergence $H$ denotes the Shannon entropy.

Now, by Pinsker's inequality, \begin{eqnarray*}\|f-N(0, I_n)\|_1 &\le& \sqrt{2 D(f\|N(0, I_n))}\\\ &=& \sqrt{\log(2\pi e)-H(f)}\end{eqnarray*}

So the conclusion is that if the entropy $f$ is closer to that of Gaussian, $f$ will be closer to $N(0, I_n)$ in total variation. But it works only for those $f$ which also has covariance matrix $I_n$.

I am not sure the following helps.

If $f$ is a prob.distribution on $\mathbb{R}^n$ having the same covariance matrix as $N(0, I_n)$, then \begin{eqnarray*}D(f\|N(0, I_n))&=& H(N(0, I_n))-H(f)\\\ &=& \frac{1}{2}\log((2\pi e)^n)-H(f)\end{eqnarray*} where $D(\cdot \| \cdot)$ denotes the KL divergence $H$ denotes the Shannon entropy.

Now, by Pinsker's inequality, \begin{eqnarray*}\|f-N(0, I_n)\|_1 &\le& \sqrt{2 D(f\|N(0, I_n))}\\\ &=& \sqrt{\log((2\pi e)^n)-2 H(f)}\end{eqnarray*}

So the conclusion is that if the entropy $f$ is closer to that of Gaussian, $f$ will be closer to $N(0, I_n)$ in total variation. But it works only for those $f$ which also has covariance matrix $I_n$.