(too long for a comment)

Let's consider the case $n=3$.

In $\mathbb P^3$, for a curve of genus $g$ and degree $d$, given a point on $x\in C$, lines through $x$ are parameterized by $\mathbb P^2$.

Lines that intersect another point in $C$ are the projection of $C$ from $x$. This is a curve of genus $g$ and degree $d-1$ in $\mathbb P^2$. (One point on this curve represents a tangent line and not a secant line).

Lines that intersect two other points in $C$ are nodes (or other multi-branch singularities) of this curves. The total number of singularities, counted by their contribution to the difference between the arithmetic and geometric genus, is $ (d-2)(d-3)/2 -g$.

In a sufficiently general situation, these will all be nodes, and so the set of lines (with a marked point) form a degree $(d-2)(d-3)/2-g$ cover of $C$. I think this is always positive unless $C$ is a plane curve (in which case this is trivial), $g=0$ and $d=3$, or $g=1$ and $d=4$.

How can we fail to be sufficiently general? Either if the projection map is generically not birational onto its image - that is if a generic line through two points on $C$ passes through a third (i.e. if the space of trisecants has dimension $2$), or if the general fiber has singularities other than nodes - which happens when a general point on $C$ lies on a tangent line (other than its own), or similar degeneracies. Probably one can eliminate these or classify when they happen, but I'm not sure.

completelinear system $\lvert L\rvert$, $S_3(C)=\varnothing$ is equivalent to $h^0(L(-D_3))=h^0(L)-3$ for all effective divisors $D_3$ of degree 3. This is the case in particular if $\deg(C)\geq 2g+2$; it is also true for the canonical embedding ($L=K$) if $C$ is not hyperelliptic or trigonal. $\endgroup$