As Ross has mentioned, you cannot know the actual value of the expressions you wrote based only on the characteristics of random variables such as mean or a variance. You can only ask for the distribution of these expressions.
E.g. in the case $\xi_n$ (rnd$_n$) are iid random variables, you can use the fact that $ \mathsf E[(1-\xi_i)(1-\xi_j)]=\mathsf E(1-\xi_i)\mathsf E(1-\xi_j) = (\mathsf E(1-\xi_i))^2$ which leads to the fact that $ \mathsf E \pi_N = \frac1N[\mathsf E(1-\xi_1)]^N = \frac{(1-a)^N}{N} $ where $a = \mathsf E\xi_1$. Here I denoted $ \pi_N = \frac{\prod\limits_{n=1}^N(1-\xi_n)}{N}. $
This holds regardless of the distribution of $\xi$, just integrability is needed. In the same way you can also easily calculate the variance of this expression based only on the variance and expectation of $\xi$ (if you want, I can also provide it).
Finally, there is a small hope that for the Gaussian r.v. $\xi$ the distribution of this expression will be nice since it includes the products of normal random variables.
On your request: variance.
Recall that for any r.v. $\eta$ holds $V \eta = \mathsf E \eta^2 - (\mathsf E\eta)^2$ hence $\mathsf E\eta^2 = V\eta+(\mathsf E\eta)^2$. As I told, you don't need to know the distribution of $\xi$, just its expectation $a$ and variance $\sigma^2$. Since we already calculated $\mathsf E\pi_N$, we just need to calculate $\mathsf E\pi^2_N$: $ \mathsf E\pi_N^2 = \frac1{N^2}\mathsf E\prod\limits_{n=1}^N(1-\xi_n)^2 = \frac{1}{N^2}\prod\limits_{n=1}^N\mathsf E(1-\xi_n)^2 = \frac{1}{N^2}\left(\mathsf E(1-\xi_1)^2\right)^N. $ Now, $ \mathsf E(1-\xi_1)^2 = \mathsf E(1-2\xi_1+\xi^2_1) = 1-2a+\mathsf E\xi_1^2 = 1-2a+a^2+\sigma^2 = (1-a)^2+\sigma^2 $ and $ \mathsf E\pi_N^2 = \frac{1}{N^2}\left((1-a)^2+\sigma^2\right)^N. $
As a consequence, $ V\pi_N = \frac{1}{N^2}\left[\left((1-a)^2+\sigma^2\right)^N - (1-a)^{2N}\right]. $