You’ve misused the multiplication rule. There are $\dfrac{n(n-1)}2$ games, and each has $2$ possible outcomes, so there are
$$\large2^{\frac{n(n-1)}2}$$
possible outcomes.
For example, with $n=3$ there are $3$ games, and each has $2$ possible outcomes, so there are $2\cdot2\cdot2=2^3$ possible results for the tournament.
There are two cases where I know how to do it, when the number of players $m$ is of the form $m=4^n$, and when $m$ is of the form $m={3^n-1\over2}$. These correspond to two types of finite geometries with $4$ points on a line. In the first case, we can consider the players as points in affine $n$ space over $\mathbb{F}_4$, the field with four elements. In the second case, we consider them as points in projective $n-1$ space over $\mathbb{F}_3.$ In either case, two points determine a line, so if we take all the lines as the games, each pair of players will meet exactly once, which is clearly the best possible. In these cases then, the minimum number of games is $$\left\lceil{m(m-1)\over12}\right\rceil.$$
I suspect these are the only cases in which this minimum can actually be achieved.
Unfortunately, these cases don't seem to cast any light on the remaining ones. In those cases, we have to have some pairs meet more than once, which eliminates any possibility of considering the games as lines.
We also need a good way of determining the minimum number is specific cases, so we can try to guess the pattern (if there is one.) For example, the best I've done so far with $8$ players is $6$ games, but is it the minimum? How to check?
P.S.
I realize of course that one can check with a computer program; I meant, "Is there an easy way to check by hand?"
EDIT
This appears to be A011976, the covering number $C(n,4,2).$ Since OEIS only lists values up to $n=20$, computation is probably hard. Apparently, it is an open problem to find a formula, but I don't know how up to-date-this site is. This paper discusses upper and lower bounds, and construction of coverings in some cases. A much more extensive table is cited by @Henry in the comments.
Best Answer
An easy way to set up such a tournament makes use of the fact that the projective plane $PG(2,3)$ consists of $13$ points on $13$ lines, each line containing $4$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).
So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $12$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.