The total number of games won must add up to $15$, since there were $15$ games. Each team must have won between $0$ and $5$ games. It's fairly obvious that the only way for none of the teams to have the same number of wins as another is if the six teams won $0,1,2,3,4$, and $5$ games.
I think this is where you made the innocent mistake. There are $6$ numbers to be distributed among the $6$ teams, which can be done in $6!=720$ different ways (not $5!=120$). The answer you seek is $\frac{720}{2^{15}}=\frac{45}{2048}$.
Label the teams as $1,2,3,4$.
The total number of ways to form the $4$ teams is
$$
n
=
\binom{28}{7}
\binom{21}{7}
\binom{14}{7}
\binom{7}{7}
$$
Let $v$ be the list of counts, sorted in ascending order, for the number of girls on each of the $4$ teams.
Consider $3$ cases . . .
Case $(1)$:$\;v=[1,1,1,4]$.
For case $(1)$, the number of ways to form the $4$ teams is
$$
x_1
=
\left(\binom{4}{1}\binom{3}{3}\right)
{\cdot}
\left(\binom{7}{4}\binom{21}{3}\right)
{\cdot}
\left(\binom{3}{1}\binom{18}{6}\right)
{\cdot}
\left(\binom{2}{1}\binom{12}{6}\right)
{\cdot}
\left(\binom{1}{1}\binom{6}{6}\right)
$$
hence the probability for case $(1)$ is
\begin{align*}
p_1&=\frac{x_1}{n}\\[4pt]
&={\Large{\frac
{
\left(\binom{4}{1}\binom{3}{3}\right)
{\cdot}
\left(\binom{7}{4}\binom{21}{3}\right)
{\cdot}
\left(\binom{3}{1}\binom{18}{6}\right)
{\cdot}
\left(\binom{2}{1}\binom{12}{6}\right)
{\cdot}
\left(\binom{1}{1}\binom{6}{6}\right)
}
{
\binom{28}{7}
\binom{21}{7}
\binom{14}{7}
\binom{7}{7}
}}}
\\[4pt]
&=\frac{2401}{59202}\approx .04055606230\\[4pt]
\end{align*}
Case $(2)$:$\;v=[1,1,2,3]$.
For case $(2)$, the number of ways to form the $4$ teams is
$$
x_2
=
\left(\binom{4}{1}\binom{3}{1}\binom{2}{2}\right)
{\cdot}
\left(\binom{7}{3}\binom{21}{4}\right)
{\cdot}
\left(\binom{4}{2}\binom{17}{5}\right)
{\cdot}
\left(\binom{2}{1}\binom{12}{6}\right)
{\cdot}
\left(\binom{1}{1}\binom{6}{6}\right)
$$
hence the probability for case $(2)$ is
\begin{align*}
p_2&=\frac{x_2}{n}\\[4pt]
&={\Large{\frac
{
\left(\binom{4}{1}\binom{3}{1}\binom{2}{2}\right)
{\cdot}
\left(\binom{7}{3}\binom{21}{4}\right)
{\cdot}
\left(\binom{4}{2}\binom{17}{5}\right)
{\cdot}
\left(\binom{2}{1}\binom{12}{6}\right)
{\cdot}
\left(\binom{1}{1}\binom{6}{6}\right)
}
{
\binom{28}{7}
\binom{21}{7}
\binom{14}{7}
\binom{7}{7}
}}}
\\[4pt]
&=\frac{2401}{6578}\approx .3650045607\\[4pt]
\end{align*}
Case $(3)$:$\;v=[1,2,2,2]$.
For case $(3)$, the number of ways to form the $4$ teams is
$$
x_3
=
\left(\binom{4}{3}\binom{1}{1}\right)
{\cdot}
\left(\binom{7}{2}\binom{21}{5}\right)
{\cdot}
\left(\binom{5}{2}\binom{16}{5}\right)
{\cdot}
\left(\binom{3}{2}\binom{11}{5}\right)
{\cdot}
\left(\binom{1}{1}\binom{6}{6}\right)
$$
hence the probability for case $(3)$ is
\begin{align*}
p_3&=\frac{x_3}{n}\\[4pt]
&={\Large{\frac
{
\left(\binom{4}{3}\binom{1}{1}\right)
{\cdot}
\left(\binom{7}{2}\binom{21}{5}\right)
{\cdot}
\left(\binom{5}{2}\binom{16}{5}\right)
{\cdot}
\left(\binom{3}{2}\binom{11}{5}\right)
{\cdot}
\left(\binom{1}{1}\binom{6}{6}\right)
}
{
\binom{28}{7}
\binom{21}{7}
\binom{14}{7}
\binom{7}{7}
}}}
\\[4pt]
&=\frac{7203}{32890}\approx .2190027364\\[4pt]
\end{align*}
Hence the probability that each team has at least one girl is
$$
p
=
p_1+p_2+p_3
=
\frac{2401}{59202}
+
\frac{2401}{6578}
+
\frac{7203}{32890}
=
\frac{16807}{26910}\approx .6245633593
$$
Best Answer
Another way: It so happens that one of the $22$ people is the Queen, who of course gets to choose the people who will be on her team. This can be done in $\binom{21}{10}$ ways.