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Considering this question, the asymptotic number of rationals in $[0,1]$ with denominator less than $n$ is $\,\sim \dfrac{3}{\pi^2}n^2$. Now in the problem we are looking at, the slopes can be any of these rationals, or their inverses, their negatives, or their negative inverses.

Therefore there are asymptotically $\dfrac{12}{\pi^2}n^2$ slopes available, and each valid configuration is using up $\dfrac{n(n-1)}{2}$ of them which represents an asymptotical proportion of $\dfrac{\pi^2}{24}$ which is roughly $41\%$. I think that part of the point lies in the fact that this proportion is uniformly bounded away from $0$.

If we consider a probabilistic approach (for what it is worth since there is clearly some bias which I will disregard): let $N\sim \dfrac{12}{\pi^2}n^2$ and consider picking successively at random $a N$ numbers between $1$ and $N$, where $0<a<1$ is a constant ($41\%$, it doesn't really matter). The probability that all these numbers are different is $$\prod_{i=1}^{aN}(1-\frac{i}{N})$$ Now to be fair we really run the experiment $n!$ times (the number of permutation matrices), and if we keep assuming independent trials, the expected number of Costas arrays of order $n$ would be $$n!\prod_{i=1}^{aN}(1-\frac{i}{N})$$ Since $N>n^2$ this still tends to $0$ pretty quickly.

Then again, this mustn't be completely right, since the works of James K. Beard seem to indicate that for higher values of $n$ the curve would get off again - see e.g. this article.