It takes $3$ comparisons to determine the median of the three elements. However, we can set the pointers to the second and second-last element and thus save two comparisons. So, we need $n+2$ comparisons before we execute quicksort on the two new arrays. Since we can assume that the three elements for the median are picked randomly, there are $k(n-k-1)$ scenarios that lead to a left array of size $k$ and a right array of size $n-k-1$. Each scenario is equally likely, thus, we have to divide by 
\begin{align}
\sum_{k=1}^{n-2}k(n-k-1)=\frac{n(n-1)(n-2)}{6}
\end{align}
The final recurrence relation describing the number of comparisons is 
\begin{align}
C_n=&n+2+\frac{6}{n(n-1)(n-2)}\sum_{k=1}^{n-2}k(n-k-1)(C_k+C_{n-k-1})\\
=&n+2+\frac{12}{n(n-1)(n-2)}\sum_{k=1}^{n-2}k(n-k-1)C_k
\end{align}
The recurrence can be simplified to 
\begin{align}
(n+1)(n+2)C_{n+2}-2(n-1)(n+1)C_{n+1}+(n + 2) (n - 5)C_n=&6 (2 n + 3)
\end{align}
Wolfram Alpha gives the solution
\begin{align}
C(n) = \frac{n + 1}{7}(c_1 +12 H_n) + \frac{720 c_2}{(n - 5) (n - 4) (n - 3) (n - 2) (n - 1) n} - \frac{54}{49} n - \frac{19}{49}
\end{align}