For $n=1$ and $d\geq 5$, we may take $m=\chi(\mathcal O(d))-3$. Take the complete intersection of two general degree $d$ curves. These points define a two-dimensional linear system. We may find among these points $\chi(\mathcal O(d))-3$ points that apply linearly independent conditions, so that they define a three-dimensional linear system.

Let's check that these $m$ points are linearly independent from any other point.

Any point that is not among the original $d^2$ points fails to lie on one of the two curves and hence provides a linearly independent condition. So we need to check that none of the other points is a base point.

We can do this using a monodromy argument. Observe that the set of base points among the original $d^2$ is determined by the choice of $m$ points. We have a canonical way to, from a choice of $m$ points in the intersection of two generic degree $d$ curves, obtain a subset of the complement. Because it is canonical, it must be invariant by the monodromy of the family of $d^2$ points. As long as the monodromy group is $S_{d^2}$, the only invariant such functions are the function that sends a subset to the empty set or to the whole complement.

Because the linear system is $3$-dimensional, there is one function in it that is not in the original linear system, which must fail to vanish at one of the remaining points. So they are not all base points. Because we know not every point in the complement is a base point, the set of base points must be empty, and we are done (conditional on the monodromy group being $S_{d^2}$).

Let's show the monodromy group is $S_{d^2}$. By the classification of $k$-transitive permutation groups for $k\geq 4$, it is sufficient to show that it is $6$-transitive. This is equivalent to saying that the expectation of, for a random element of the monodromy group, the number ordered $6$-tuples of fixed points, is $1$.

We can compute this expectation using Deligne's equidistribution theorem. One consequence of this is that for a variety $X$ over a finite field $\mathbb F_q$ and a map from $\pi_1(X)$ to a finite group that is surjective on the geometric fundamental group, the distribution of Frobenius elements of a random point in $X(\mathbb F_q)$ is close to the distribution of random elements of $G$, and the distribution gets closer as $q$ grows.

Here we take $X$ to be the space of pairs of degree $d$ plane curves that intersect transversely and the map to be the map to the monodromy group as a subgroup of $S_{d^2}$. We need to know that the expectation of the number of ordered $6$-tuples of $\mathbb F_p$-points of a complete intersection of two random degree $d$ curves over $\mathbb F_p$ is $1+o(1)$. This is true by counting: The number of $6$-tuples of points of $\mathbb P^2$ is $\frac{(p^2+p+1)!}{(p^2+p-5)!}$. Because $d\geq 5$, the points apply linearly independent conditions, so the probability of such a tuple lying on the complete intersection of two random degree $d$ curves is $1/p^{12}$. Hence the total expectation is:

$$\frac{(p^2+p+1)!}{p^{12}(p^2+p-5)!}=1+o(1)$$

This monodromy computation is due to Alexei Entin.

There is an alternative interpretation of the same arugment that does not use finite fields. The complete intersections form a $d^2$-fold cover of the space of pairs of degree $d$ curves with transverse intersection. The set of $6$-tuples of points in this cover forms a $\frac{ (d^2)!}{(d^2-6)!}$-fold cover. We want to show that the monodromy group acts transitively on this cover. Equivalently, we want to show that it is irreducible. It is sufficient to show that the space of pairs of two degree $d$ curves and $6$ distinct points on both curves is irreducible, without the transverse intersection condition. We can do this by viewing it as a bundle over the space of $6$-uples of points on $\mathbb P^2$, which is irreducible. The fibers of this bundle are the spaces of pairs of degree $d$ curves through $6$ distinct points, which are a product of projective spaces. Because the conditions imposed by the points are linearly independent, the projective spaces form a fiber bundle. Because we have an irreducible bundle on an irreducible base, the scheme is irreducible.