Suppose there are $n$ variables. If we could easily integrate probabilities over all $n$ variables at once, we would find the exact value for $p$, and would have no need of bounds. Instead we will provide bounds by a single two-variable integral.
We go from $r=n$ to $r=2$, where
\begin{align}
B^n=0,\text{ and generally }&B^r \text{ is }n\times 1\ \text{ and constant }\\
M^n=1,\text{ and generally }&M^r \text{ is }n\times r\ \text{ and constant }\\
Y^n=Z,\text{ and generally }&Y^r \text{ is }r\times 1\ \text{ and random }\\
\lambda^n=\mu,\text{ and generally }&\lambda^r \text{ is }r \times 1\ \text{ and the mean of }Y,\\
\Pi^r=\Sigma,\text{ and generally }&\Pi^r \text{ is }r\times r\ \text{ and the covariance of }Y,
\end{align}
At each step we define
$$p=P[Y^r_1=\max M^rY^r+B^r, Y^r \sim N(\lambda^r,\Pi^r)]$$
(we omit superscripts on scalars). We will get upper [lower] bounds by choosing variables so that the $p$'s increase [decrease] as $r$ decreases.
Let $\theta^r_{ij} := \arccos(\Pi^r_{kij}/\sqrt{\Sigma_{ii}\Sigma_{jj}}) = \arccos(\text{cor}(Y^r_i,Y^r_j))$
Let $\ell$ be such that $\lambda^r_\ell=\min_{i>1}\lambda^r_i$.
Let $k$ be such that
$\lambda^r_k=\min_{i>1,i\neq \ell}\lambda^r_i$.
Let $h$ be the number distinct from $k$ and $\ell$ which maximizes $\cos(\theta_{hk}+\theta_{h\ell})$. (For lower bounds, find $h$ which maximizes $\cos(|\theta_{hk}-\theta_{h\ell}|)$ instead.)
Let $T$ be the same as $\Sigma$ except that $T_{k\ell}=\sigma_k \sigma_\ell \cos(\theta_{hk}+\theta_{h\ell}) \le \Sigma_{k\ell}$
$T$ has the lowest possible correlation between $Y_k$ and $Y_\ell$ given their correlations with $Y_h$, so it describes a distribution where $Y_h$, $Y_k$, and $Y_\ell$ are linearly dependent.
Let $a,b,c$ be such that $Y_\ell = a Y_h + b Y_k + c$ under $T$.
Let $B^{r-1} = B^r + cM^r_{\cdot k}$.
Let $M^{r-1}_i$ (the ith row of the matrix) be the same as $M^r_i$ but with
$M^{r-1}_{ih}=M^r_{ih}+aM^r_{i\ell}$,
$M^{r-1}_{ik}=M^r_{ik}+bM^r_{i\ell}$ and $M^{r-1}_{i\ell}$ omitted.
Let $\lambda^{r-1}$ be $\lambda^r$ but with the entry for $Z_\ell$ omitted.
Let $\Pi^{r-1}$ be $T$ but with the row and column for $Z_\ell$ omitted.
Then the new $p$ will be larger than the old $p$, as desired. $\square$