Suppose \begin{align*} \begin{pmatrix} X_i \\ Y_i \\ Z_i \end{pmatrix} \sim N\left(\begin{pmatrix} \mu_X \\ \mu_Y \\ \mu_Z \end{pmatrix}, \sigma^2\begin{pmatrix} 1 & \rho_{XY} & \rho_{XZ} \\ \rho_{XY} & 1 & \rho_{YZ} \\ \rho_{XZ} & \rho_{YZ} & 1 \end{pmatrix}\right) \end{align*} The sample correlation estimator \begin{align*} \widehat{\rho}_{XY} = \frac{\sum_{i=1}^{n}X_iY_i - n \overline{X}\overline{Y}}{(n-1)S_XS_Y}, \qquad S_X = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(X_i - \overline{X})^2} \end{align*} is known to have approximate distribution \begin{align*} \text{arctanh}(\widehat{\rho}_{XY}) \overset{\mathcal{D}}{\approx} N\left(\text{arctanh}(\rho_{XY}), \frac{1}{\sqrt{n-3}}\right) \end{align*} or, more correctly, \begin{align*} \sqrt{n}[\text{arctanh}(\widehat{\rho}_{XY})-\text{arctanh}(\rho_{XY})] \overset{\mathcal{D}}{\rightarrow} N\left(0, 1\right) \end{align*} and an analogous statement can be made about $\widehat{\rho}_{YZ}$. From there, if we really wanted, we can apply the delta method to derive the asymptotic distributions of $\widehat{\rho}_{XY}, \widehat{\rho}_{YZ}$.
In fact, we can slightly rewrite the previous results into the form \begin{align*} \sqrt{n}\begin{pmatrix} \text{arctanh}(\widehat{\rho}_{XY}) - \text{arctanh}(\rho_{XY}) \\ \text{arctanh}(\widehat{\rho}_{YZ}) - \text{arctanh}(\rho_{YZ}) \ \end{pmatrix}\overset{\mathcal{D}}{\rightarrow} N\left(\begin{pmatrix} 0 \\ 0\end{pmatrix}, \begin{pmatrix} 1 & a \\ a & 1\end{pmatrix}\right) \end{align*} My Question: What is $a$? That is, the asymptotic covariance between $\widehat{\rho}_{XY}$ and $\widehat{\rho}_{YZ}$?