Here's a relatively easy counterexample. Take $p=2$, $L=\mathbb{Q}_2(2^{1/3})$. I did some direct computation and saw that $v_2(\mathfrak{D}^{L_2}_L)=2/3$, $v_2(\mathfrak{D}^{L_3}_{L_2})=5/6$. Looks like there's a pattern. But there's a better way.

We're in a situation where not only $L/\mathbb{Q}_2$ is tamely ramified of degree $3$, but also every $L_n/K_n$. Now we use the functoriality of the Hasse-Herbrand transition function: if $k\subset F\subset K$, then $\varphi^K_k=\varphi^F_k\circ \varphi^K_F$. Use the relation $\varphi^{L_n}_{\mathbb{Q}_2}=\varphi^{K_n}_{\mathbb{Q}_2}\circ \varphi^{L_n}_{K_n}= \varphi^L_{\mathbb{Q}_2}\circ \varphi^{L_n}_ L$ and the fact that a tamely ramified extension has all the transition-function action at the origin. That is, the function is $y=x$ for $x\le 0$, but $y=x/e$ for $x\ge 0$, where $e$ is the ramification index. So as real functions, $\varphi^L_{\mathbb{Q}_2}=\varphi^{L_n}_{K_n}$, namely this is just $y=x/3$. Consequently, the transition function of $L_n/L$ is gotten by conjugating that of $K_n/{\mathbb{Q}_2}$ with the tame transition function. The effect is to multiply all coordinates of vertex points by $3$.

But we also know the transition function of $K_n$ over ${\mathbb{Q}_2}$: its vertices are at all $(2^{i-1}-1,i-1)$ for $2\le i\le n$.The new vertices are at $(3,3)$, $(9,6)$, $(21,9)$, etc. This means that the lower breaks of $L_n/L$ are at $3(2^{i-1}-1)$ for $2\le i\le n$, and in particular the unique break of $L_n/L_{n-1}$ is at $3(2^{n-1}-1)$.

Now use the formula
\begin{align}{
v_F(\mathfrak{D}^F_k)=\sum_{j\ge 0}\bigl(|G_j|-1\bigr)
}\end{align}
where the $G$'s are the lower ramification groups, and where in this case all the numbers being added up are $1$ or $0$, to see that $v_{L_n}\bigl(\mathfrak{D}^{L_n} _ {L_{n-1}}\bigr)=3(2^{n-1}-1)+1=3\cdot2^{n-1}-2$. Divide by the ramification number of $L_n$ over $\mathbb{Q}_2$ to get $1-1/(3\cdot 2^{n-2})$, agreeing with my computations for $n=2$ and $n=3$.

It's not an issue of tame versus wild ramification in the extension $L/\mathbb{Q}_2$, either. I used $L=\mathbb{Q}_2(2^{1/4})$ to find that the numbers are $1-3/2^m$; the argument is similar but a bit more delicate, since you have no a priori idea of what the transition function of $L_n/K_n$ might be.