Let's solve a simpler problem first: How many binary search trees are there with $n$ nodes? Call this number $b_n$. We know $b_0 = b_1 = 1$. If you pick $k$ as the root ($1 \le k \le n$), you have $k - 1$ keys to distribute to the left and $n - k - 1$ to the right, so that:
$$
b_n = \sum_{1 \le k \le n} (b_{k - 1} + b_{n - k - 1})
$$
Shift the indices down, and note the symmetry of the sum to get:
$$
b_{n + 1} = 2 \sum_{0 \le k \le n} b_k
$$
Define the generating function $B(z) = \sum_{n \ge 0} b_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\sum_{n \ge 0} b_{n + 1} z^n
= 2 \sum_{n \ge 0} z^n \sum_{0 \le k \le n} b_k
$$
Recognize some sums here:
$$
\frac{B(z) - b_0}{z}
= 2 \frac{B(z)}{1 - z}
$$
Pugging in $b_0 = 1$ solution is:
$$
B(z) = \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{1 - 3 z}
$$
so that (here $[\text{condition}]$ is Iverson's bracket, $1$ if the condition is true and $0$ if false):
$$
b_n = \frac{2 \cdot 3^n + [n = 0]}{3}
$$
Now we are in position to solve your problem if there is no restriction on height. If the root of a binary search tree with $n$ nodes is $k$, there are $k - 1$ nodes in its left subtree and $n - k$ to the right, so the number of binary trees with root $k$ are $b_{k - 1} \cdot b_{n - k}$, and you are asking for:
$\begin{align}
\sum_{1 \le k \le n} k b_{k - 1} b_{n - k}
&= \sum_{1 \le k \le n}
k \frac{2 \cdot 3^{k - 1} + [k = 1]}{3}
\cdot \frac{2 \cdot 3^{n - k} + [k = n]}{3} \\
&= 1 \cdot \frac{2 \cdot 3^0 + 1}{3} \cdot \frac{2 \cdot 3^{n - 1}}{3}
+ n \cdot \frac{2 \cdot 3^{n - 1}}{3} \cdot \frac{2 \cdot 3^0 + 1}{3}
+ 4 \cdot 3^{n - 3} \sum_{2 \le k \le n - 1} k \\
&= (1 + n) \cdot 2 \cdot 3^{n - 2}
+ 4 \cdot 3^{n - 3} \left( \frac{n (n - 1)}{2} - 1 \right) \\
&= (2 n^2 + 4 n + 2) \cdot 3^{n - 3} \\
&= 2 (n + 1)^2 \cdot 3^{n - 3}
\end{align}$