Feller was able to determine the median asymptotically, namely $N \log_2 N$, or $\log_2 N$ per game. He actually proved an even stronger concentration bound. This result can also be found in his "Introduction to Probability".
Economists are still coming up with new interpretations of this problem, known as the St. Petersburg Paradox, after the city in which Euler, who invented it, was working (the name might be due to Feller). See, for example, this article, which also suggests the same value.
From this paper we have the inequality
$\frac{x}{\log(x)}(1 + \frac{1}{\log(x)} + \frac{2}{\log^2(x)}) < \pi(x) < \frac{x}{\log(x)}(1 + \frac{1}{\log(x)} + \frac{2.334}{\log^2(x)})$ for $x \ge 2 953 652 287$
which leads to $2.347112 \cdot 10^{102} < p_{10^{100}} < 2.347127 \cdot 10^{102}$.
EDIT: I just noticed that the same paper gives
$$p_k < k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2}{\log k}\right)$$ for $k \ge 688383$ and
$$p_k > k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2.1}{\log k}\right)$$ for $k \ge 3$ which yields $2.3471221 \cdot 10^{102} < p_{10^{100}} < 2.3471265 \cdot 10^{102}$, so six digits are determined.
EDIT 2: Thanks to DanaJ, I see that this 2013 paper by Axler gives the following bounds:
$p_k < k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2}{\log k}\right) - \frac{(\log(\log(k)))^2 - 6 \log(\log(k)) + 11.847}{(\log (k))^2}$ for $k \ge 2$ and
$p_k > k \left(\log(k) + \log(\log(k)) - 1 + \frac{\log(\log(k))-2}{\log k}\right) - \frac{(\log(\log(k)))^2 - 6 \log(\log(k)) + 10.273}{(\log (k))^2}$ for $k \ge 8009824$
which yields $2.347125652 \cdot 10^{102} < p_{10^{100}} < 2.347125801 \cdot 10^{102}$, determining the first seven digits.
Note that, while this paper gives the better asymptotic bound
$|\pi(x) - \mathrm{li}(x)| < 0.2795\frac{x}{(\log (x))^{3/4}}\exp(-\sqrt{\frac{\log (x)}{6.455}})$ for $x \ge 229$
it only determines the first three digits of $p_{10^{100}}$.
Of course, if we assume the Riemann Hypothesis we can get many more digits. The bound
$|\pi(x) - \mathrm{li}(x)| < \frac{\sqrt{x} \log(x)}{8\pi}$ for $x \ge 2657$
will give
$$2.347125735865764178036135909936302071965422425975\cdot10^{102}<p_{10^{100}}<2.347125735865764178036135909936302071965422425983*10^{102}$$
so $47$ digits of $p_{10^{100}}$ are determined.
Best Answer
The first thing you have to notice is that it is a round robin.
Looking at an easier example, with two teams, each team will play against the other once, how many games will there be? Assuming teams are $A$ and $B$, then there is $1$ game in total: $A$ vs $B$
If there are three, then there is $AvB, AvC, BvC$ so $3$
If there are four, then there is $AvB, AvC, AvD, BvC,BvD, CvD$, so 6.
You might notice a pattern here - assuming the number of teams is $n$, the first team plays $n-1$ games. The second plays $n-2$ games (they actually play $n-1$, but we already counted one of those games, the one against the first team)
Therefore it's the sum from $1 \to n-1$, which is ${(n-1 + 1)(n-1) \over 2} = \frac{n*(n-1)}{2}$ At then, the answer is then $45$