First, as the Quora answer makes clear, you can think of this problem simply as the ants moving without colliding at all, since whether they collide or not, the end effect is still having two ants moving in the two different directions.
Second, with $100$ ants spread out and having $1$ cm between them, the two outmost ants start $0.5$ cm from the ends, and so the upper bound is $99.5$ seconds, while the two most inside ants start $49.5$ cm from the ends, and so the lower bound is $49.5$ seconds.
Third, the probability distribution for the possible outcomes will be a geometric distribution. Indeed, the expected time will be very close to the upper bound.
To see this, consider: if at least one of the outer ants start by walking inward, it'll take $99.5$ seconds, and the chance of that is pretty good: $\frac{3}{4}$.
And even if both outside ants immediately walk off (with a chance of $\frac{1}{4}$), then there is a $\frac{3}{4}$ chance that at least one of the second to the most outer ants start walking inward, in which case it'll take $98.5$ seconds. So, there is a $\frac{3}{16}$ chance it will take $98.5$ seconds.
And we get exponentially smaller chances for that time to go down, e.g. there is a $\frac{3}{64}$ chance it will take $97.5$ seconds, a $\frac{3}{256}$ chance it will take $96.5$ seconds, etc. Like I said, you get a geometric distribution with smaller and smaller probabilities. As you found yourself, the probability of getting the lower bound is $1$ in $2^{100}$. ut even getting something below $90$ is highly improbable, as you'd need all of the $20$ outmost ants face the right way, so that only happens $1$ in $2^{20}$ times.
Finally, here is my rough calculation for the expected time ... someone who is better with series (especially partial sums) should be able to give a more precise answer ...
$$E = \sum_{i=0}^{49} (\frac{3}{4})\cdot (\frac{1}{4})^{i} \cdot (99.5-i) + (\frac{1}{4})^{50} \cdot 49.5 \approx$$
$$\sum_{i=0}^{\infty} (\frac{3}{4})\cdot (\frac{1}{4})^{i} \cdot (99.5-i) = $$
$$\frac{3}{4}\cdot \big(99.5 \cdot \sum_{i=0}^{\infty} (\frac{1}{4})^{i} -\sum_{i=0}^{\infty} (\frac{1}{4})^{i} \cdot i \big) = $$
$$\frac{3}{4}\cdot \big(99.5 \cdot \frac{1}{1-\frac{1}{4}} -\frac{1}{4} \cdot \sum_{i=1}^{\infty} (\frac{1}{4})^{i-1} \cdot i \big) = $$
$$\frac{3}{4}\cdot \big(99.5 \cdot \frac{1}{\frac{3}{4}} -\frac{1}{4} \cdot \frac{1}{(1-\frac{1}{4})^2}\big) = $$
$$\frac{3}{4}\cdot \big(99.5 \cdot \frac{4}{3} -\frac{1}{4} \cdot \frac{16}{9}\big) = $$
$$99.5 -\frac{3}{9} \approx 98.8333 $$
Note that if you use $101$ ants, so that the two outer ants are at the very end of the stick, and the middle ant is in the middle, you get an upper bound of $100$ seconds, a lower bound of $50$ seconds, and an expected time of a little over $99$ seconds.
