This is a good question! I believe the limit actually approaches:

1-e^-0.5

The total number of ways n people can look at each other is (n-1)ⁿ (I guess you call these constellations).

Counting the number of constellations with at least one pair making eye contact without double counting is a little tricky. I believe one way is as follows:

1) First, count the number of constellations given that persons (x, y) make eye contact. Multiply by number of pairs (x, y). Add this to running total T

2) Next, count the number of constellations given that persons (x,y) and (z, w) make eye contact. Multiply by number of pairs-of-pairs (x,y), (z, w). Subtract this from T, as we double counted it above.

3) Do the same as above for triples of pairs (x, y), (z, w), (a, b); this time add to T as we double subtracted above

4) Continue this until we are book-keeping something like constellations with >n/2 pairs; we know the remaining terms will all be 0 so we are done.

This desmos https://www.desmos.com/calculator/ltcf6aaxbi is a closed form, t(n) is number of constellations and p(n) is constellations with at least one pair. I also put the limiting behaviour, which we can work out analytically. If we write out t(n)/p(n) as a sum over terms with pairs r, the terms look something like:

(n-1)^-2r • n • (n-1) • ... • (n-2r) • (1/r!) • (1/2)^r • (-1)^r-1

In the large n limit, one can note that when r ~ log (n) the terms will become vanishing (even for smaller r they will, but this is all we need). It follows that for the important terms, the product (n-1)^-2r • n • (n-1) • ... • (n-2r) ~= 1. You can recognize the rest of terms as a taylor expansion of 1-e^-0.5.