I'm not much of an expert on Brill-Noether Theory, so apologies if this is is a silly question. Given $p_1,\ldots, p_n$ distinct points in $\mathbb P^{n-1}$, where $n\geq 4$, is it possible to find at least $n+1$ linearly independent quadrics that pass through $p_1,\ldots, p_n$. If this is not true in general, are there specific instances where this is true?
For example, if $v_1,\ldots, v_n$ are an orthogonal basis for $\mathbb K^n$ (where $\mathbb K$ need not be algebraically closed), then if $\ell_i = v_i\cdot [x_1,\ldots, x_n]^T$, the quadrics $\ell_i\ell_j$ are linearly independent and all pass through $v_1,\ldots, v_n$. As $n\geq 4$, there are at least $n+1$ linearly independent quadrics passing through $v_1,\ldots, v_n$.
This is true. Here's one way to see this if your points are in general position, i.e. if I let $l_i \subset k^n$ to be the line corresponding to the point $p_i$, then I'm assuming that taking $0\ne v_i\in l_i$, you have that $\{v_i\}$ are linearly independent in $k^n$.
1) WLOG you may assume that your points are given by $p_1 = [1:0:\ldots:0]$, $p_2=[0:1:\ldots:0], \ldots, p_n = [0:0:\ldots:0:1]$. To see this, because $\{v_i\}$ are linearly independent, there exists $A\in GL(n)$ such that $Av_i = \langle e_i \rangle$, where $\{e_i\}$ are your standard basis of $k^n$. Then it follows that $A$ defines an automorphism of $\mathbb{P}^{n-1}$ mapping $p_1$ to $[1:0:\ldots:0]$ etc.
2) Let $X_0,\ldots,X_{n-1}$ denote homogeneous coordinates on $\mathbb{P}^{n-1}$. A quadric can be written as $\sum_{i=0}^{n-1} a_{ii} X_i^2 + \sum_{i\ne j} a_{ij} X_iX_j = 0$, where $a_{ij} \in k$. Imposing the conditions that they vanish on $p_i$ for all $i$, we deduce that $a_{ii} = 0$ for all $i$. This leaves us that the vector space of quadrics vanishing on $p_i$ for all $i$ has dimension $\binom{n}{2} \ge n+1$ for $n\ge 4$.
If your points are not in general position, i.e. if the $\{v_i\}$ are not linearly independent, then basically tracing through the same argument will find that the vector space of quadrics is going to have even higher dimension.
Another way of saying this is simply to note that you have an exact sequence
$$0\rightarrow \mathcal{I}(2) \rightarrow \mathcal{O}_{\mathbb{P}^{n-1}}(2) \rightarrow \mathcal{O}_{\bigcup p_i} \rightarrow 0$$
Applying the long exact sequence in cohomology and noting that $h^0(\mathcal{O}_{\mathbb{P}^{n-1}}(2)) = \binom{n+1}{2}$ and $h^0(\mathcal{O}_{\bigcup p_i}) = n$ tells you that $h^0(\mathcal{I}(2)) \ge \binom{n-1}{2} \ge n+1$ when $n\ge 4$, which is the same claim as above.