In the following I am arguing that the result is due to the law of large numbers and the convergence of the sum of Laplace distributed random variables towards a normal distribution.
Ansatz
Let's keep it simple and assume - without loss of generality - a univariate setting. As in your question text, we define the the draw down $DD$ as the largest peak-to-trough movement along an observed time series $S_i$ of length $N$, $S_1, S_2,\ldots S_N$, i.e.
$$
\begin{align}
DD_N &\equiv \max_{i,j}S_i-S_j \\
\mathrm{s. t.} &\quad 1\leq i<j\leq N
\end{align}
$$
As in your question, given two points $i<j$ in our series, we introduce the cumulative (log) return $R(i,j)\equiv \sum_{k=i+1}^jr_k$ and identify
$$
S_j\equiv S_ie^{R(i,j)}
$$
with $r_k\equiv\ln(S_k)-\ln(S_{k-1})$ the $k$th log return observation.
Statement 1: Longer (observation) time periods will result in larger price variation, and hence in larger drawdown (statistically):
Given the setup above we find that a move from $i$ to $j$ (which could be the drawdown) is
$$
\begin{align}
S_j-S_i&=S_ie^{R(i,j)}-S_i\\
&=S_i\left(e^{R(i,j)}-1\right)\\
&=S_i\left(e^{\sum_{k=i+1}^jr_k}-1\right)
\end{align}
$$
Hence the drawdown $DD_N$ is (of course) directly linked to the cumulative sum of individual return contributions. Somewhat informally: the larger the potential period our measure covers ($j-i+1$ will increase in probability when $N$ increases), the larger the potential for a large drawdown is. Again, informally
$$
DD_N > DD_M \quad \forall N>M
$$
This should be no news to you: The longer the time horizon $j-i+1$ during which we allow for (bad) individual returns to accumulate by letting $N$ grow, the larger the potential drawdown can be.
Statement 2: Longer cumulative return series will converge to a normal distribution
Yet, as $N$ increases, and under proper technical conditions, the distribution of the sum of $N$ (sufficiently) independent random variables will converge to that of a normal distribution by the central limit theorem. Thus, for a sufficiently large number $N$ in your example, the maximum draw down sampled from Laplace distributed returns will converge to that of a normal distribution with appropriately chosen parameters.
We can see that this is indeed the case if we have a look at the kurtosis of a sum of $N$ Laplace distributed random variables centered around zero. The moment generating function of the Laplace distribution is
$$
M_X(t)\equiv \mathrm{E}\left(e^{tX}\right)=\frac{1}{1-b^2t^2}
$$
for some dispersion parameter $b$. Likewise, the MGF of the sum of $N$ independent Laplace trials is
$$
M_{Z_N=\sum_{i=1}^NX_i}(t)\equiv \mathrm{E}\left(e^{t\sum_{i=1}^NX_i}\right)=\frac{1}{\left(1-b^2t^2\right)^N}
$$
From the same wiki page, we know how to calculate the first four moments of the distribution using the moment generating function and find
$$
\mathrm{E}(X)=0\quad\mathrm{E}(X^2)=2b^2\quad\mathrm{E}(X^3)=0\quad\mathrm{E}(X^4)=24b^4\quad
$$
and
$$
\mathrm{E}(Z_N)=0\quad\mathrm{E}(Z_N^2)=2b^2N\quad\mathrm{E}(X^3)=0\quad\mathrm{E}(X^4)=12b^4N(N+1)\quad
$$
And we can now show that the kurtosis of $Z_N$ will converge to that of a normally distributed random variable as $N\to \infty$
$$
\begin{align}
\mathrm{Kurt}(Z_N)&\equiv \frac{\mathrm{E}\left(\left(Z-\mathrm{E}(Z)\right)^4\right)}{\mathrm{E}\left(\left(Z-\mathrm{E}(Z)\right)^2\right)^2}\\
&=\frac{\mathrm{E}\left(Z^4\right)}{\mathrm{E}\left(Z^2\right)^2}\\
&=\frac{12b^4N(N+1)}{4b^4N^2}\\
&=3\left(1+\frac{1}{N}\right)
\end{align}
$$
Supporting simulation study
Please find below the results of a corresponding simulation study conducted using R. I have based the simulation on n trading days worth of simulated daily returns (using the normal and the Laplace distribution). For convenience, I am assuming a volatility of 20%, a mean return of nil, return series of length $N=2, 20, 50, 252$ observations. I have stuck to your setup of nSim = 50000 simulated paths.
NB: I have chosen the parameters of the Laplace distribution such that the simulated return volatility is indeed 20% and I have adjusted the simulated series such that the expected value is $E(S_j|S_i)=S_i$ under both setups.
nSim <- 50000
sigma <- 0.2
dt <- 1 / 252
maxDD <- function(z){
dd <- 0.0
for (i in 1:(length(z)-1)){
cand <- max(z[i]-z[-(1:i)])
if (cand>dd){dd <- cand}
}
dd
}
n <- 2
normal <- normal <- sapply(1:nSim,function(i){
maxDD(exp(c(0,cumsum(
-0.5 * sigma^2 * dt + sigma*sqrt(dt)*rnorm(n=n)
))))})
laplace <- sapply(1:nSim,function(i){
maxDD(exp(c(0,cumsum(
-log(2) * sigma^2 * dt + sigma*sqrt(dt)*ExtDist::rLaplace(n=n,mu = 0,b=sqrt(2)/2)
))))})
qqplot(normal,laplace)
abline(a=0,b=1)
title(main=paste("N = ",n))
Finally, I have created QQ-plots of the resulting Drawdown distributions for each hypothetical series length below. As we can see from the diagrams, the simulated drawdown distributions converge as per your observation. For small $N$, the tails of the Laplace have an effect over the normal distribution, but as soon as we simply 'allow' for more returns to enter the drawdown, the influence of the central limit theorem outweighs the tails fo the Laplace distribution.
