Suppose there are $ N $ random variables $X_i\sim N(\mu_i,\sigma^2) $, where you only know $\sigma^2$. You do not know the underlying distribution of each $X_i $, and instead have a prior over $\mu_i $, which is $\mu_i\sim N(\mu_0,\sigma_0^2) $. You observe one realization of $ X_i$ for each $i $, and update your prior for each $i $, giving $ N $ posteriors $ \mu_i\mid X_i\sim N(\mu_i', \sigma'^{2})$. What is the probability distribution of the maximum order statistic of the updated posterior means $\mu_i'$?
I know that we have $$\mu_i\mid X_i \sim N\left(\frac{X_i\sigma_0^2+\mu_0\sigma^2}{\sigma_0^2+\sigma^2},\frac{\sigma_0^2\sigma^2}{\sigma_0^2+\sigma^2}\right)\implies \mu_i'= \frac{X_i\sigma_0^2+\mu_0\sigma^2}{\sigma_0^2+\sigma^2}.$$ and we may write
$$ \mu_i\sim N(\mu_0,\sigma_0^2)\implies \mu_i=\mu_0+Z\sigma\implies X_i\sim N(\mu_0+Z\sigma_0,\sigma)\implies X_i=\mu_0+Z\sigma_0+Z\sigma, $$ where $Z\sim N(0,1)$. Does it suffice to simply write
\begin{align*}
\mu_i\mid X_i&= \frac{X_i\sigma_0^2+\mu_0\sigma^2}{\sigma_0^2+\sigma^2}+Z\left(\frac{\sigma_0\sigma}{\sqrt{\sigma_0^2+\sigma^2}}\right)\\
&= \frac{(\mu_0+Z\sigma_0+Z\sigma)\sigma_0^2+\mu_0\sigma^2}{\sigma_0^2+\sigma^2}+Z\left(\frac{\sigma_0\sigma}{\sqrt{\sigma_0^2+\sigma^2}}\right)\\
&=\frac{\mu_0\sigma_0^2+\mu_0\sigma^2}{\sigma_0^2+\sigma^2}+Z\left(\frac{\sigma_0^3}{\sigma_0^2+\sigma^2}\right)+Z\left(\frac{\sigma\sigma_0^2}{\sigma_0^2+\sigma^2}\right)+Z\left(\frac{\sigma_0\sigma}{\sqrt{\sigma_0^2+\sigma^2}}\right)\\&=\underbrace{\mu_0+Z\left(\frac{\sigma_0^3}{\sigma_0^2+\sigma^2}\right)+Z\left(\frac{\sigma\sigma_0^2}{\sigma_0^2+\sigma^2}\right)}_{\text{mean, } \mu_i'}+Z\left(\frac{\sigma_0\sigma}{\sqrt{\sigma_0^2+\sigma^2}}\right)\\\mu_i'&\sim N\left(\mu_0,\frac{\sigma_0^4}{\sigma_0^2+\sigma^2}\right)
\end{align*}
And then use that normal distribution in the canned formulas for order statistics?
You wrote $X_i\sim N(\mu_i, \sigma^2).$ I would write $X_i\mid \mu_i\sim N(\mu_i,\sigma^2)$ so as to distinguish between this and the marginal distribution of $X_i.$
You wrote $\mu_i\sim N(\mu_0,\sigma_0^2)$ but did not say that the $\mu_i$ are independent for all the different values of $i,$ but I will assume that that was intended.
You wrote $X_i = \mu_0 + Z\sigma_0 + Z\sigma.$ That is correct if the two $Z\text{s}$ are independent of each other. I might have called them $Z_1$ and $Z_2$ or something like that. Apparently you continue to follow that convention below that.
Since $\mu_i,\,i=1,\ldots,n$ are independent, so are $\operatorname E(\mu_i\mid X_i) = \dfrac{X_i\sigma_0^2+\mu_0\sigma^2}{\sigma_0^2+\sigma^2}, \, i=1,\ldots,n.$ Therefore $\mu_i,\,i=1,\ldots,n$ make up an i.i.d. sample from a certain normal distribution, which you found, so the "canned formulas for order statistics" will work.