I'd take this approach:
Call $ p_d= P(x_1 > x_2 + d)$
The probability that sample $x_1$ is the largest value AND it exceeds the second largest in more than $d$ is just $p_d^{N-1}$
The probability that the largest value exceeds the second largest in more that $d$ is then $N p_d^{N-1}$ WRONG-fixed below
To compute $p_d$: that is the probability that the difference of two iid normals exceeds $d$. But the difference of two normals is a normal of media cero, and variance $2 \sigma^2$, so that probability is given by the integral of the normal cumulative distribution function. What matters to us is that it's a constant that don't depend on $N$.
So the probability in question (fixed $d$, $N \to \infty$) tends to zero.
UPDATE: As correctly pointed out in the comments, the second step is wrong, the events are not independent. They are independent, though, if $x_1$ is fixed - that is, they are conditionally independent. So:
$P(x_1 > x_2 + d | x_1) = F_x(x_1 - d) $
$P(x_1 > x_i + d | x_1) = F_x^{N-1}(x_1 - d)$
And $P(x_1 > x_i + d) = \int_{-\infty}^{\infty}F_x^{N-1}(x - d) f_x(x) dx$
(this tends to $1/N$ as $d \to 0^+$, as was to be expected)
So, restricting to $d \ge 0$, the probability that the largest value exceeds the second largest in more that $d$ is (if I have not messed up anything else) :
$ P(x_A - x_B > d)= N \int_{-\infty}^{\infty} F_x^{N-1}(x - d) f_x(x) dx $
where $x_A$ is the largest value and $x_B$ the second one.
The question is about the above formula going to zero or not for fixed $d$, and growing N. That seems to depends on the density.
As a check: if $x$ is exponential with parameter $\lambda$, the integrals can be evaluated, and I get:
$ P(x_A - x_B > d) = exp( - \lambda d) $
This (besides tending to 1 for $d \to 0^+$, as it should) does not depend on $N$. Hence, for an exponential distribution the two events of the original question are equally surprising.