Spanning tree in a graph of intersecting sets
Interesting question.
The right intuition should probably be along the guideline that two random subsets of cardinality $n$ drawn from some $cn$ elements for some constant $c$ differ from each other significantly with a probability very close to 1 and, hence, the weight of the minimum spanning tree of the graph $G$ should be $\mathcal\Theta(n^2)$ on average. I cannot prove that guideline is correct, however.
Instead, I will present one series of such examples. More specifically, from some $n$ (that can be arbitrary large), there are $n$ sets, each having $(n-1)/2$ elements drawn from a set of $n$ elements, such that the cardinality of the symmetric difference between any two sets is no less than $(n-1)/2$. So the weight of the minimum spanning tree is no less than $(n-1)^2/2=\mathcal\Theta(n^2)$.

Here is the construction, using quadratic residue.
Example. Let $n=p$ be an odd prime. Let $X_0$ be the set of all non-zero quadratic residues of $p$ between 0 and $p-1$ inclusive. In other words, $$X_0=\{0\le k\lt p: \left(\frac {k}p\right)=1\}$$ where $\left(\frac{\cdot}p\right)$ is the Legendre symbol. For $0\le i\lt p$, let $X_i$ be &quot;$X_0$ plus $i$&quot;, i.e., $$X_i=\{0\le k\lt p: \left(\frac {k-i}p\right)=1\}.$$ Then $|X_i|=\frac{p-1}2$ for all $i$ and $|X_i \triangle X_j|\ge \frac{p-1}2$ for all $i\not=j$.
Proof: Since $\left(\frac{\cdot}p\right)$ is either $-1$, $0$, or $1$, we have $1+\left(\frac{\cdot}p\right)\ge0$. Hence,
$$\begin{aligned}
&amp;\quad\quad \sum_{0\le k\lt p}\left(1+\left(\frac {k-i}p\right)\right)\left(1+\left(\frac {k-j}p\right)\right)\\
&amp;\ge\sum_{0\le k\lt p\,\land\,\left(\frac {k-i}p\right)=1\,\land\, \left(\frac {k-j}p\right)=1}\left(1+\left(\frac {k-i}p\right)\right)\left(1+\left(\frac {k-j}p\right)\right)\\
&amp;=\sum_{0\le k\lt p\,\land\,\left(\frac {k-i}p\right)=1\,\land\, \left(\frac {k-j}p\right)=1}4\\
&amp;=4\,|X_i\cap X_j|
\end{aligned}$$
On the other hand, we have
$$\begin{aligned}
&amp;\quad\quad \sum_{0\le k\lt p}\left(1+\left(\frac {k-i}p\right)\right)\left(1+\left(\frac {k-j}p\right)\right)\\
&amp;=\sum_{0\le k\lt p}\left(1 + \left(\frac {k-i}p\right) + \left(\frac {k-j}p\right)+ \left(\frac {k-i}p\right)\left(\frac {k-j}p\right)\right)\\
&amp;=p + 0 + 0 + \sum_{0\le k\lt p} \frac {k^2-(i+j)k+ij}p\\
&amp;=p-1
\end{aligned}$$
Since $p\nmid(-(i+j))^2-4ij=(i-j)^2$, the last equality above comes from the case of $p\nmid b^2-4ac$, theorem 1 in the paper On Certain Sums with Quadratic Expressions Involving the Legendre Symbol. So we have $|X_i\cap X_j|\le \frac{p-1}4.$
Since $|X_i|=|X_j|=\frac{p-1}2$, $\ |X_i \triangle X_j|=|X_i|+|X_j|-2|X_i\cap X_j|\ge \frac{p-1}2.$ $\quad\checkmark$


For people who appreciate concrete examples, here are the sets when $n=17$, where $|X_i \triangle X_j|\ge 8$.
$$\begin{aligned}
X_{0}&amp;=\{\phantom{1}1, \phantom{1}2, \phantom{1}4, \phantom{1}8, \phantom{1}9, 13, 15, 16 \}\\
X_{1}&amp;=\{\phantom{1}2, \phantom{1}3, \phantom{1}5, \phantom{1}9, 10, 14, 16, \phantom{1}0 \}\\
X_{2}&amp;=\{\phantom{1}3, \phantom{1}4, \phantom{1}6, 10, 11, 15, \phantom{1}0, \phantom{1}1 \}\\
X_{3}&amp;=\{\phantom{1}4, \phantom{1}5, \phantom{1}7, 11, 12, 16, \phantom{1}1, \phantom{1}2 \}\\
X_{4}&amp;=\{\phantom{1}5, \phantom{1}6, \phantom{1}8, 12, 13, \phantom{1}0, \phantom{1}2, \phantom{1}3 \}\\
X_{5}&amp;=\{\phantom{1}6, \phantom{1}7, \phantom{1}9, 13, 14, \phantom{1}1, \phantom{1}3, \phantom{1}4 \}\\
X_{6}&amp;=\{\phantom{1}7, \phantom{1}8, 10, 14, 15, \phantom{1}2, \phantom{1}4, \phantom{1}5 \}\\
X_{7}&amp;=\{\phantom{1}8, \phantom{1}9, 11, 15, 16, \phantom{1}3, \phantom{1}5, \phantom{1}6 \}\\
X_{8}&amp;=\{\phantom{1}9, 10, 12, 16, \phantom{1}0, \phantom{1}4, \phantom{1}6, \phantom{1}7 \}\\
X_{9}&amp;=\{10, 11, 13, \phantom{1}0, \phantom{1}1, \phantom{1}5, \phantom{1}7, \phantom{1}8 \}\\
X_{10}&amp;=\{11, 12, 14, \phantom{1}1, \phantom{1}2, \phantom{1}6, \phantom{1}8, \phantom{1}9 \}\\
X_{11}&amp;=\{12, 13, 15, \phantom{1}2, \phantom{1}3, \phantom{1}7, \phantom{1}9, 10 \}\\
X_{12}&amp;=\{13, 14, 16, \phantom{1}3, \phantom{1}4, \phantom{1}8, 10, 11 \}\\
X_{13}&amp;=\{14, 15, \phantom{1}0, \phantom{1}4, \phantom{1}5, \phantom{1}9, 11, 12 \}\\
X_{14}&amp;=\{15, 16, \phantom{1}1, \phantom{1}5, \phantom{1}6, 10, 12, 13 \}\\
X_{15}&amp;=\{16, \phantom{1}0, \phantom{1}2, \phantom{1}6, \phantom{1}7, 11, 13, 14 \}\\
X_{16}&amp;=\{\phantom{1}0, \phantom{1}1, \phantom{1}3, \phantom{1}7, \phantom{1}8, 12, 14, 15 \}\\
\end{aligned}$$