We have $N_1$, $N_2$, $N_3$ normally distributed random variables with three different means $µ_i$, but all with the same standard deviation $\sigma=1$.
I want to know the probability that $N_1$ is the largest of the three random variables.
I can solve this for two random variables. In this case I create a new random variable $d_{1,2}$ with $d_{1,2} = N_1 - N_2$. $d_{1,2}$ is normally distributed with mean $µ_1 - µ_2$ and standard deviation $\sqrt 2$. It is easy to calculate the portion of the distribution of $d_{1,2}$ that is larger than $0$. And if $d_{1,2} > 0$, this is equivalent to $N_1 > N_2$. Example: $µ_1 = \sqrt 2$, $µ_2 = 0$, $p(d_{1,2} > 0) = \Phi(1) = 0.84$.
I have not yet found a solution for three normally distributed random variables.
You have to specify the joint distribution. I gather from your question that you are additionally given that the three normal random variables are independent.
If so, the question can be framed as - what is the probability that $N_1 > N_2 \wedge N_3$. The answer to that question is a simply integral of the joint density function $f(N_1, N_2, N_3)=f_1(N_1)f_2(N_2) f_3(N_3)$ over a certain region.
$$\int_{N_1=N_2}^{\infty}\int_{N_2=N_3}^{\infty}\int_{N_3=-\infty}^{\infty} f_1f_2f_3 dN_3dN_2dN_1 +\int_{N_1=N_3}^{\infty}\int_{N_2=-\infty}^{N_3}\int_{N_3=-\infty}^{\infty} f_1f_2f_3 dN_3dN_2dN_1$$ where the expressions for the integrands are known.