Given $T$ observed vectors $x_i\in\mathbb{R}^N, i\in\{1,\ldots,T\}$. Define $\hat{\Sigma}$ as the corresponding empirical covariance-matrix of the Observations $X=\left(\begin{array}{c} x_1' \\ \vdots\\ x_T' \end{array}\right) \in\mathbb{R}^{T \times N}$.
\begin{align}
\hat{\Sigma}:=\frac{1}{T-1}X'(I_T-\frac{1}{T}\iota_T\iota_T')X
\end{align}
I am searching for an intuitive representation of $ \hat{\Sigma}^{-1}$.
Based on this answer, if the inverse of $X'X$ exists, one idea is to rewrite $$\hat{\Sigma}^{-1}=(T-1)\left(X'X-\underbrace{\frac{1}{\sqrt{T}}X'\iota_T}_{:=B}\underbrace{\iota_T'X\frac{1}{\sqrt{T}}}_{B'}\right)^{-1}\\=(T-1)\left((X'X)^{-1}-\frac{1}{1+B'(X'X)^{-1}B}(X'X)^{-1}BB'(X'X)^{-1}\right)\\ =(T-1)\left((X'X)^{-1}-\underbrace{\frac{1}{T+\iota_T'X(X'X)^{-1}X'\iota_T}}_{:=c}(X'X)^{-1}X'\iota_T\iota_T'X(X'X)^{-1}\right)\\ =(T-1)(X'X)^{-1}\left((X'X)-\frac{1}{c}X'\iota_T\iota_T'X\right)(X'X)^{-1} $$ If we assume the number of observations $T$ is high, the scalar $c:=T+\iota_T'X(X'X)^{-1}X'\iota_T\approx T.$ Especially as $\iota_T'X(X'X)^{-1}X'\iota_T>0$ we know $\frac{1}{c}<\frac{1}{T}$. Given we are satisfied with this approximation we can write $$ \hat{\Sigma}^{-1}\approx (T-1)(X'X)^{-1}\left((X'X)-\frac{1}{T}X'\iota_T\iota_T'X\right)(X'X)^{-1}\\ =(T-1)^2(X'X)^{-1}\hat{\Sigma}(X'X)^{-1} $$
