We may assume that $I$ is maximal with $D(I) >1$. If $I$ is maximal and $V(I)$ is not contained in a line, then $V(I)$ is just three points in a triangle and and $I$ is the ideal of those $3$ points. If $I$ is contained on a line, then it either has two points, one with non-reduced structure pushing slightly off the line $(x,y^2)(x-1,y)$ or has one double point $(x^2,xy,y^2)$.
Now in each case we can do this fairly explicitly. For $3$ points, $I^n$ is the ideal of functions vanishing of order $n$ at those $3$ points. This contains a function of degree $(3/2)n$, which is the product of powers of the lines through the points.
This is optimal, because given a polynomial $f$, which is the first line raised to the power $a$ times a polynomial of degree $d-a$, the polynomial of degree $d-a$ must intersect the two points on the first line with multiplicity $n-a$, so $d-a \geq 2(n-a)$ and if $d \leq (3/2) n$, $a \geq n/2$. Then the same is true for the multiplicity of the other $3$ lines, hence $d \geq 3n/2$.
For one point with multiplicity $1$ and one point with multiplicity $2$, we can again manage $(3/2)n$ , with the polynomial $y^n x^{n/2}$. Conversely if the degree is at most $(3/2)n$ the intersection multiplicity with the line $x$ is at least $2n$ so by the same logic the polynomial contains a factor of $x^{n/2}$. The remainder of the polynomial must vanish to order $n$ at $(1,0)$ hence have degree at least $n$, so the minimum is $(3/2)n$.
For the last case, $I^n$ is a homogeneous ideal generated by everything with degree above $2n$.
So we see if $D(I^n) < (3/2)n$, then $D(I)=1$.