Here's something which shows the constructions yielding $2n$ are almost tight.  

Let $A$ be the $|U| \times n$ matrix where the entry $a_{ij}$ is equal to $1$ if $i \in S_j$ and $-1$ otherwise, and let $B=A^T A$.    

Then $B$ is an $n \times n$ matrix having diagonal entries equal to $|U|$ and off-diagonal entries equal to 
$$b_{ij}=|U| - 2 (|S_j \cap S_i^C| + |S_i \cap S_j^C|)$$
$$=|U|-4(n- |S_i \cap S_j| ) \leq |U| - 2n.$$

Letting $x$ be the $n \times 1$ vector of $1$'s, we have by direct computation
$$(Ax)^T(Ax) =n (|U| -2n)^2$$
Conversely, by the above we have 
$$(Ax)^T (Ax) = \sum_i \sum_j b_{ij} = n |U| + \sum_{(i,j), i \neq j} b_{ij} \leq n|U| + n(n-1) (|U|-2n)$$

Comparing, we have $(|U|-2n)^2 \leq |U|+(n-1) (|U|-2n)$.  Since $|U|$ is an integer, this implies $|U| \geq 2n-1$ (if $|U| \leq 2n-2$ the right side is non-positive while the left side is positive).