My answer to the second question you list has a simple form of the more general result given by @whuber, but is readily adapted to the general case. Instead of $$P(X_1 > \max X_i \mid X_1 = \alpha) = \prod_{i=2}^n P\{X_i < \alpha \mid X_1 = \alpha\} = \left[\Phi(\alpha)\right]^{n-1}$$ which applies when the $X_i$ are independent $N(0,1)$ random variables, we have $$P(X_1 > \max X_i \mid X_1 = \alpha) = \prod_{i=2}^n P\{X_i < \alpha \mid X_1 = \alpha\} = \prod_{i=2}^n \Phi\left(\frac{\alpha-\mu_i}{\sigma_i}\right)$$ since the $X_i$ are independent $N(\mu_i, \sigma_i^2)$ random variables, and instead of $$P(X_1 > \max X_i) = \int_{-\infty}^{\infty}\left[\Phi(\alpha)\right]^{n-1} \phi(\alpha-\mu)\,\mathrm d\alpha$$ we have $$P(X_1 > \max X_i) = \int_{-\infty}^{\infty}\prod_{i=2}^n \Phi\left(\frac{\alpha-\mu_i}{\sigma_i}\right) \frac{1}{\sigma}\phi\left(\frac{\alpha-\mu_1}{\sigma_1}\right)\,\mathrm d\alpha$$ where $\Phi(\cdot)$ and $\phi(\cdot)$ are the cumulative distribution function and probability density function of the standard normal random variable. This is just whuber's answer expressed in different notation.
The complementary probability $P(X_1 < \max X_i) = P\{(X_1 < X_2) \cup \cdots \cup (X_1 < X_n)$ can also be bounded above by the union bound discussed in my answer to the other question. We have that $$\begin{align*} P(X_1 < \max X_i) &= P\{(X_1 < X_2) \cup \cdots \cup (X_1 < X_n)\\ &\leq \sum_{i=2}^n P(X_1 < X_i)\\ &= \sum_{i=2}^n Q\left(\frac{\mu_1 - \mu_i}{\sqrt{\sigma_1^2 + \sigma_i^2}}\right) \end{align*}$$ since $X_i-X_1 \sim N(\mu_i-\mu_1,\sigma_i^2+\sigma_1^2)$. Note that $Q(x) = 1-\Phi(x)$ is the complementary standard normal distribution function. The union bound is very tight when $\mu_1 \gg \max \mu_i$ and the variances are roughly comparable even for large $n$, but for small $n$, the bound can exceed $1$ and thus be useless.