For any tournament $T$ satisfying the problem condition, denote by $S(T)$ the sum under consideration, namely
$$S(T)=\sum_{i=1}^{n}\left(w_{i}-\ell_{i}\right)^{3}$$
First, we show that the statement holds if a tournament $T$ has only 4 players. Actually, let $A=\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ be the number of wins of the players; we may assume that $a_{1} \geq a_{2} \geq a_{3} \geq a_{4}$. We have $a_{1}+a_{2}+a_{3}+a_{4}=\binom{4}{2}=6$, hence $a_{4} \leq 1$. If $a_{4}=0$, then we cannot have $a_{1}=a_{2}=a_{3}=2$, otherwise the company of all players is bad. Hence we should have $A=(3,2,1,0)$, and $S(T)=3^{3}+1^{3}+(-1)^{3}+(-3)^{3}=0$. On the other hand, if $a_{4}=1$, then only two possibilities, $A=(3,1,1,1)$ and $A=(2,2,1,1)$ can take place. In the former case we have $S(T)=3^{3}+3 \cdot(-2)^{3}>0$, while in the latter one $S(T)=1^{3}+1^{3}+(-1)^{3}+(-1)^{3}=0$, as desired.

Now we turn to the general problem. Consider a tournament $T$ with no bad companies and enumerate the players by the numbers from 1 to $n$. For every 4 players $i_{1}, i_{2}, i_{3}, i_{4}$ consider a "sub-tournament" $T_{i_{1} i_{2} i_{3} i_{4}}$ consisting of only these players and the games which they performed with each other. By the abovementioned, we have $S\left(T_{i_{1} i_{2} i_{3} i_{4}}\right) \geq 0$. Our aim is to prove that
$$S(T)=\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S\left(T_{i_{1} i_{2} i_{3} i_{4}}\right)$$
where the sum is taken over all 4-tuples of distinct numbers from the set $\{1, \ldots, n\}$. This way the problem statement will be established.

We interpret the number $\left(w_{i}-\ell_{i}\right)^{3}$ as following. For $i \neq j$, let $\varepsilon_{i j}=1$ if the $i$-th player wins against the $j$-th one, and $\varepsilon_{i j}=-1$ otherwise. Then
$$\left(w_{i}-\ell_{i}\right)^{3}=\left(\sum_{j \neq i} \varepsilon_{i j}\right)^{3}=\sum_{j_{1}, j_{2}, j_{3} \neq i} \varepsilon_{i j_{1}} \varepsilon_{i j_{2}} \varepsilon_{i j_{3}}.$$
Hence,
$$S(T)=\sum_{i \notin\left\{j_{1}, j_{2}, j_{3}\right\}} \varepsilon_{i j_{1}} \varepsilon_{i j_{2}} \varepsilon_{i j_{3}}.$$
To simplify this expression, consider all the terms in this sum where two indices are equal. If, for instance, $j_{1}=j_{2}$, then the term contains $\varepsilon_{i j_{1}}^{2}=1$, so we can replace this term by $\varepsilon_{i j_{3}}$. Make such replacements for each such term; obviously, after this change each term of the form $\varepsilon_{i j_{3}}$ will appear $P(T)$ times, hence
$$S(T)=\sum_{\left|\left\{i, j_{1}, j_{2}, j_{3}\right\}\right|=4} \varepsilon_{i j_{1}} \varepsilon_{i j_{2}} \varepsilon_{i j_{3}}+P(T) \sum_{i \neq j} \varepsilon_{i j}=S_{1}(T)+P(T) S_{2}(T).$$
We show that $S_{2}(T)=0$ and hence $S(T)=S_{1}(T)$ for each tournament. Actually, note that $\varepsilon_{i j}=-\varepsilon_{j i}$, and the whole sum can be split into such pairs. Since the sum in each pair is 0, so is $S_{2}(T)$.

Thus the desired equality (2) rewrites as
$$S_{1}(T)=\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S_{1}\left(T_{i_{1} i_{2} i_{3} i_{4}}\right).$$
Now, if all the numbers $j_{1}, j_{2}, j_{3}$ are distinct, then the set $\left\{i, j_{1}, j_{2}, j_{3}\right\}$ is contained in exactly one 4-tuple, hence the term $\varepsilon_{i j_{1}} \varepsilon_{i j_{2}} \varepsilon_{i j_{3}}$ appears in the right-hand part of (3) exactly once, as well as in the left-hand part. Clearly, there are no other terms in both parts, so the equality is established.