Let $v\sim N(0,\sigma_v^2)$, let $z, y_1, \dots, y_n\sim N(v, \sigma^2)$ be i.i.d stochastic variables.
Calculate $E[v\mid z, z \geq y_i \ \forall i \in \{1,\dots , n\}]$
Said in another way, then there are $n+1$ agents who all have a noisy signal about the value $v$. The player with the highest signal, knows his signal and he knows that he has the highest signal. What is the expected value of $v$, conditioned on his information? Is it necessary for him to know the true distribution of $v$?
According to the properties of conditional expectation, we have: $$\begin{align} \mathbb{E}(v|z,\{y_i\le z\}_{i=1,...,n}) &= \frac{\mathbb{E}(v\mathbf{1}_{\{\{y_i\le z\}_{i=1,...,n}\}}|z) }{\mathbb{P}(\{y_i\le z\}_{i=1,...,n}|z)}\\ &= \frac{\mathbb{E}(\mathbb{E}(v\mathbf{1}_{\{\{y_i\le z\}_{i=1,...,n}\}}|z,v)|z) }{\mathbb{E}(\mathbf{1}_{\{y_i\le z\}_{i=1,...,n}}|z)}\\ &= \frac{\mathbb{E}(\mathbb{E}(v\mathbf{1}_{\{\{y_i\le z\}_{i=1,...,n}\}}|z,v)|z) }{\mathbb{E}(\mathbb{E}(\mathbf{1}_{\{y_i\le z\}_{i=1,...,n}}|z,v)|z)}\tag{1} \end{align}$$ The numerator of $(1)$: $$\begin{align} N&:=\mathbb{E}(\mathbb{E}(v\mathbf{1}_{\{\{y_i\le z\}_{i=1,...,n}\}}|z,v)|z)\\ &=\mathbb{E}(v\mathbb{E}(\mathbf{1}_{\{\{y_i\le z\}_{i=1,...,n}\}}|z,v)|z)\\ &=\mathbb{E}\left(v\left(\int_{-\infty}^z\frac{e^{-\frac{(t-v)^2}{2\sigma^2}}}{\sqrt{2\pi}\sigma}dt \right)^n|z\right)\\ &=\int_{u\in\mathbb{R}}\left(u\left(\int_{-\infty}^z\frac{e^{-\frac{(t-u)^2}{2\sigma^2}}}{\sqrt{2\pi}\sigma}dt \right)^n\varphi\left({\frac{u+z}{\sigma_v}}\right)\right)du\\ \end{align}$$ where $\varphi(\cdot)$ is the density function of $\mathcal{N}(0,1)$. Indeed, the density function of $v|z$ is determined as follows $$z\stackrel{\mathcal{D}}{=} v+\sigma\mathcal{N}(0,1)\implies v|z \stackrel{\mathcal{D}}{=} z - \sigma\mathcal{N}(0,1)\implies f_{v|z}(u)=\varphi\left(\frac{u+z}{\sigma_v} \right)$$
A simpler representation of $N$ is $$N = \int_{u\in\mathbb{R}}u\Phi^n\left(\frac{z-u}{\sigma}\right)\varphi\left({\frac{u+z}{\sigma_v}}\right)du$$ where $\Phi(\cdot)$ is the probability function of $\mathcal{N}(0,1)$.
By similar technique, the denominator of $(1)$ is equal to: $$D:=\mathbb{E}(\mathbb{E}(\mathbf{1}_{\{y_i\le z\}_{i=1,...,n}}|z,v)|z)=\int_{u\in\mathbb{R}}\Phi^n\left(\frac{z-u}{\sigma}\right)\varphi\left({\frac{u+z}{\sigma_v}}\right)du$$ We conclude that: $$\color{red}{\mathbb{E}(v|z,\{y_i\le z\}_{i=1,...,n}) = \frac{\int_{u\in\mathbb{R}}u\Phi^n\left(\frac{z-u}{\sigma}\right)\varphi\left({\frac{u+z}{\sigma_v}}\right)du}{\int_{u\in\mathbb{R}}\Phi^n\left(\frac{z-u}{\sigma}\right)\varphi\left({\frac{u+z}{\sigma_v}}\right)du}}$$