For the Trapezoidal Rule, you actually use $n+1$ points. For example, in the simple case where you are integrating $f(x)$ from $0$ to $1$, and you want $T_4$, you evaluate $f$ at the points $0/4$, $1/4$, $2/4$, $3/4$, and $4/4$. It is $n+1$ points because we use the endpoints.
For the Midpoint Rule, you use $n$ points, but these are not the same points as for the Trapezoidal Rule. They are the midpoints of our intervals. So in the example discussed above, for $M_4$ you would be evaluating $f$ at $1/8$, $3/8$, $5/8$, and $7/8$.
The Simpson Rule $S_{2n}$ uses evaluation of $f$ at $2n+1$ points. If for example $n=4$, then you are dividing the interval into $8$ subintervals. With $n=4$ and the interval $[0,1]$, you would be using the points $0/8$, $1/8$, $2/8$, $3/8$, $4/8$, $5/8$, $6/8$, $7/8$, and $8/8$.
Note that $1/8$, $3/8$, $5/8$ and $7/8$ are the points that were used for $M_4$. The points $0/8$, $2/8$, $4/8$, $6/8$, and $8/8$ are just $0/4$, $1/4$, $2/4$, $3/4$, and $4/4$, exactly the points that were used for $T_4$.
A more abstract summary: $T_n$ uses $n+1$ points, and $M_n$ uses $n$ points. But the $n$ points used by $M_n$ are completely different from the points used for $T_n$. So altogether, $T_n$ and $M_n$ carry information about function evaluation at $2n+1$ points, which is exactly what $S_{2n}$ does.
I have not written out a proof of the formula, only tried to deal with your discomfort with the $2n$ on one side and $n$'s on the other. The formula is not hard to verify. Let's do it explicitly for $n=4$. Write down, say for the interval $[0,1]$, what $T_4$ is.
We have
$$T_4=\frac{1}{8}(f(0)+2f(1/4)+2f(1/2)+2f(3/4)+f(1)).$$
Now write down $M_4$:
$$M_4=\frac{1}{4}(f(1/8)+f(3/8)+f(5/8)+f(7/8)).$$
Now calculate $T_4+2M_4$. It is convenient for the addition to make sure that $M_4$ has denominator $8$, so write $2M_4$ as $\frac{1}{8}(4f(1/8)+4f(3/8)+4f(5/8)+4f(7/8))$, and add. Divide by $3$ and you will get the expression you would get in $S_{8}$. The same method works in general.
Best Answer
Note that "these formulas" give only an error estimate. The actual error may be much smaller.
In order to obtain an example where the formula for $E_S$ produces a larger error than the formula for $E_T$ consider the function $$f(x):=\sin(\omega x)\qquad(0\leq x\leq1)\ .$$ You then have $|f''(x)|\leq\omega^2$ and $|f^{(4)}(x)|\leq \omega^4$. This gives $$|E_T|\leq {\omega^2\over12 n^2},\qquad |E_S|\leq{\omega^4\over180 n^4}\ ,$$ so that the estimate for $E_S$ is worse than the estimate for $E_T$ as soon as $\omega^2\geq 15 n^2$.