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Some lower bounds:

In the case where $I$ is radical, if $D(I) \geq 2$, then $D(I^n) \geq (3/2)n$ (and in fact $\lceil (3/2) n \rceil$ is achieved.) $V(I)$ must not be contained in any line, so it must contain $3$ noncolinear points, and we can assume that $I$ is the ideal of $3$ noncolinear points. Then $I^n$ 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 $d\geq 3n/2$.

Here's another interesting phenomenon. Take $I$ to be the ideal of $k (k+1) /2$ generic points. Then $D(I)= k$ by dimension counting. $I^n$ is the ideal of functions vanishing of order $n$ at $k(k+1)/2$ distinct points, which is an ideal of codimension $n (n+1)/2 \cdot k (k+1)/2$. This is less than $d (d+1)/2$ for $d$ approximately equal to $nk / \sqrt{2}$. So there is a degree $d$ polynomial in $I^n$, and $D(I^n)$ is asymptotically at most $nk/\sqrt{2}$.

I can show that if $D(I) \geq 2$, then $\lim_{n \to \infty} D(I^n)/ n> 1$. Take $I$ maximal with respect to the property $D(I) \geq 2$. Then each local factor of $I$ at a point of $V(I)$ either contains two linear functions, or is maximal with respect to the property of containing one linear function, and hence looks like $(y, x^2)$, or is maximal with respect to the property of containing no linear functions, and hence looks.

What do the last kind of ideals look like? There must be some length $1$ extension, which must contain some linear function $y$, and so it is of the form $(x^k,y)$ for some $n$. Length one extensions of that have the form $(x^{k+1}, xy, y^2, ax^k+ by)$ and we must have $a \neq 0$. If $b =0$, the ideal contains is contained in $(x^2, xy, y^2)$, which is one example of a maximal ideal with this property. Otherwise by scaling $y$, we may put it in the form of my example.

Case 1: $I= (x^2, xy, y^2)$. An element in $I^n$ vanishes to order $2n$ on $I$, hence has degree at least $2n$.

Case 2: $I = (y-x^k, x^{k+1})$. An element in $I^n$ intersects $y-x^k$ with multiplicity $n (k+1)$, hence has degree at least $n (k+1)/k$. Having $(y-x^k)$ divide the element doesn't help because it has degree $k$ but only removes $k+1$ of the intersection.

Case 3: $I$ contained in $(y, x^2)$. Then $I$ must also vanish somewhere else on the line $y=0$. If the degree is at most $(3/2)n$ the intersection multiplicity with the line $x=0$ is at least $2n$ so by the same logic as in the reduced case the polynomial contains a factor of $x^{n/2}$. The remainder of the polynomial must vanish to order $n$ at the other point on the line $y=0$ hence have degree at least $n$, so the minimum is $(3/2)n$.

Case 4: $I$ is contained in none of these and is maximal, hence reducd. We already did this case.

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 $(3/2)n$ when $n$ is even, 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 $(3/2)n$ when $n$ is even, 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 $D(I)=1$.

EDIT: I made a mistake in the second case and the answer is actually no. Suppose $V(I)$ is two points. $I$ must be the product of the ideals of the two points. Wlog they are $(0,0)$ and $(1,0)$. If the ideals of both points contain the function $y$, then by chinese remainder theorem the product contains $y$. Wlog $(0,0)$ does not contain $y$. It's maximal with this property, so a length one extension of it does contain $y$, hence is of the form $(x^k,y)$ for some $n$. Length one extensions of that have the form $(x^{k+1}, xy, y^2, ax^k+ by)$ and we must have $a \neq 0$. If $b =0$, the ideal contains $(x^k, xy, y^2)(x-y,y)$, which is contained in the ideal I described.

Otherwise by scaling $y$ and removing unnecessary equations we may write it as $(y-x^k, x^{k+1})(x-1,y)$.

Unfortunately this ideal to the power of $k$ includes $y^{k+1}$, because:

$$ y^{k+1} = (y-x^k + x^k)^{k+1} = \sum_{i=0}^{k+1} \begin{pmatrix} k+1 \\ i \end{pmatrix} (y-x^k)^i x^{k (k+1-i) } $$

Because $k (k+1)- i \geq (k+1) (k-i)$, we lose.

Consider $I = (y- x^k, x^{k+1})$.

For $k>1$ this does not contain any linear functions. It contains $xy$ so $D(I)=2$. But I claim $y^{k+1} \in I^k$, so $D(I^k) = k+1$.

By the binomial theorem

$$ y^{k+1} = (y-x^k + x^k)^{k+1} = \sum_{i=0}^{k+1} \begin{pmatrix} k+1 \\ i \end{pmatrix} \left(y-x^k\right)^i x^{k (k+1-i) } $$

In the exponent:

$$k(k+1-i) =k^2 +k - ik = (k+1)(k-i) + i$$

so this is

$$(y-x^k)^{k+1} + \sum_{i=0}^{k} \begin{pmatrix} k+1 \\ i \end{pmatrix} x^i\left(y-x^k\right)^i \left(x^{k+1}\right)^{k-i} \in I^k$$

(Boris pointed out a flaw in my earlier argument, leading me to find this counterexample.)

In general, subadditivity shows $\lim_{n \to \infty} \frac{D(I^n)}{n}$ exists, and that any fixed value of $\frac{D(I^n)}{n}$ is at least this limit. So one version of this question is about how to compare $D(I)$ to this limit. Here we show the limit can go arbitrarily close to $1$ with $D(I)=2$. By adding random linear factors, the limit can get arbitrarily close to $D(I)-1$. But probably for larger $D(I)$ the limit can be less than $D(I)$ by even more than $1$.

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 $D(I)=1$.

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 $(3/2)n$ when $n$ is even, 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 $(3/2)n$ when $n$ is even, 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 $D(I)=1$.

EDIT: I made a mistake in the second case and the answer is actually no. Suppose $V(I)$ is two points. $I$ must be the product of the ideals of the two points. Wlog they are $(0,0)$ and $(1,0)$. If the ideals of both points contain the function $y$, then by chinese remainder theorem the product contains $y$. Wlog $(0,0)$ does not contain $y$. It's maximal with this property, so a length one extension of it does contain $y$, hence is of the form $(x^k,y)$ for some $n$. Length one extensions of that have the form $(x^{k+1}, xy, y^2, ax^k+ by)$ and we must have $a \neq 0$. If $b =0$, the ideal contains $(x^k, xy, y^2)(x-y,y)$, which is contained in the ideal I described.

Otherwise by scaling $y$ and removing unnecessary equations we may write it as $(y-x^k, x^{k+1})(x-1,y)$.

Unfortunately this ideal to the power of $k$ includes $y^{k+1}$, because:

$$ y^{k+1} = (y-x^k + x^k)^{k+1} = \sum_{i=0}^{k+1} \begin{pmatrix} k+1 \\ i \end{pmatrix} (y-x^k)^i x^{k (k+1-i) } $$

Because $k (k+1)- i \geq (k+1) (k-i)$, we lose.