To ease the statement, I use the notation for real vectors and matrices. As $\epsilon_i = \boldsymbol{e}_i^T \boldsymbol{\epsilon}$, where $\boldsymbol{e}_i \in \mathbb{R}^N$(Sometimes called the $i$-th standard basis vector) with all elements equal to 0 expect the $i$-th element equal to 1, we have $$ E(\epsilon_i^2) = E[(\boldsymbol{e}_i^T (I - G) \boldsymbol{x})^2] = \boldsymbol{e}_i^T (I - G)(I - G^T) \boldsymbol{e}_i$$ or $$ E(\epsilon_i^2) = \|(I - G^T) \boldsymbol{e}_i\|_2^2. $$ BTW, the given condition $E\{\boldsymbol{x}^H\boldsymbol{x}\} = I$ should be $E\{\boldsymbol{x}\boldsymbol{x}^H\} = I$.

To simplify the statement, I will use the following notation for real vectors and matrices. Let $\epsilon_i = \boldsymbol{e}_i^T \boldsymbol{\epsilon}$, where $\boldsymbol{e}_i \in \mathbb{R}^N$. Sometimes $\boldsymbol{e}_i$ is called the $i$-th standard basis vector; it has all elements equal to 0 except the $i$-th element equal to 1. We have $$ E(\epsilon_i^2) = E[(\boldsymbol{e}_i^T (I - G) \boldsymbol{x})^2] = \boldsymbol{e}_i^T (I - G)(I - G^T) \boldsymbol{e}_i$$ or $$ E(\epsilon_i^2) = \|(I - G^T) \boldsymbol{e}_i\|_2^2. $$ By the way, the given condition $E\{\boldsymbol{x}^H\boldsymbol{x}\} = I$ should be $E\{\boldsymbol{x}\boldsymbol{x}^H\} = I$.