The effect of ignoring covariance on variance estimates Suppose we fit a generalized linear mixed model of the form
$$y_{ij} = x_{ij}^T \beta + u_i$$
and under the model it is assumed that $u_i \sim N_p (0,\Sigma)$ where $\Sigma$ is assumed to be a diagonal matrix with diagonal elements $\sigma_1^2,\dots,\sigma_p^2$.
Suppose in reality that $u_i$ comes from a multivariate normal distribution where $\Sigma$ is not diagonal, so that correlation exists between the random terms in $u_i$.
I'm interested in how this sitation would affect our estimates of $\sigma_1^2,\dots,\sigma_p^2$.
Intuitively, I think that if they are positively correlated this would cause us to underestimate $\sigma_1^2,\dots,\sigma_p^2$ since the data possess less variation than that implied by the model with diagonal matrix. Conversely, if $\sigma_1^2,\dots,\sigma_p^2$ possess some kind of negative correlation, I think the reverse might be true, depending on the exact structure.
I'm hoping someone could explain how correlation between  random intercepts would affect our estimates under the assumption of independence in a more formal/statistical way
 A: This is not a general answer but only one example:
Assume that $(X_1,X_2) \sim N_2(\mu,\Sigma)$ with  $\mu=(0,0)^T$. 
Furthermore assume that $X_2=aX_1+X_3$, where $X_3 \sim N(0,\sigma^2)$ and $X_1$ and $X_3$ are independent. Then we have positive correlation between $X_1$ and $X_2$ if $a>0$ and negative correlation if $a<0$.
In this case wrongly assuming in our model that we have a normal with only diagonal elements (i.e.  $X_1$ and $X_2$ are independent ) corresponds to assuming that $a=0$, while in reality e.g. $a>0$. 
This means that the model that we are actually interested in (as we assume that the components in our model are independent) is $(X_1,X_2-aX_1)=(X_1,X_3) \sim N_2(\mu,\Sigma_T)$, where $$\Sigma_T = \begin{bmatrix}Var(X_1) & 0 \\ 0& Var(X_3) \end{bmatrix}$$
However what we actually end up doing is to estimate $Var(X_2)=Var(aX_1+X_3)=a^2Var(X_1)+Var(X_3)$ (due to our wrong model) .
However (as I understand your question) you are interested in only the Variance of $X_3$, therefore you would overestimate the variance.
A more intuitive example: Assume that you want to measure the variance of the shots of a soccer player on a certain target and the variance of the wind and you assume that they are independent. However it turns out that the wind also affects the variance of the shots of the soccer player. If you are only interested in the variance of the soccer player that is due to him, the assumption of independence leads you to overestimate his variance, as part of it is due to the wind and not due to him.
