I've thought about this for a bit, and I now believe that leadtime demand in most common situations is not normally distributed, although it may be as usual a good approximation.
Of course, we know that the normal distribution has infinite tails which means you could argue that it is not ever appropriate for non-negative random variables, like demand. This is a silly argument; after all, the Central Limit Theorem holds that the distribution of the sum of a large number of independent non-negative random variables tends to a normal distribution.
When modeling demand over time, I find it useful to think about a cumulative stochastic process $D(t)$ counting all demand from time zero to $t$ where $D(0)=0$. If this process has independent increments and if the distribution of $D(t)$ is assumed to be normal with mean $\mu_D t$ and variance $\sigma_D^2 t$, then it is a classic Brownian motion with drift coefficient $\mu$ (also assuming continuous paths in time, which is somewhat technical).
After some thought, I have realized that while $D(t)$ is a normal random variable for any fixed $t$, it is not true in general that $D(L)$ is a normal random variable for a stochastic lead time $L$. The simplest counterargument is to suppose that $L$ is a Bernoulli random variable; in this case, the probability density of $D$ would include a point mass of $p$ at zero mixed with a continuous density everywhere else. This is not any normal density of course.
The last important question is how good is a normal approximation in this case, with mean $\mu_D\Bbb E[L]$ and variance $\sigma_D^2 \Bbb E[L] + \mu^2 \operatorname{var}(L)$, for setting an appropriate reorder point.
Edit: Ideas after @LarrySnyder610 answer
Larry's answer shows that when we assume $L$ is nearly normally-distributed, the distribution of $D(L)$ also appears to be well-approximated by a normal distribution. In his answer, $L$ is a truncated non-negative normal random variable, taking value $0$ with probability $P(N(\mu_L, \sigma^2_L)\leq 0)$ but having a normal density for $L > 0$. When $\mu_L$ is large enough and $\sigma^2_L$ is small enough, my guess is that the normal approximation is a good one.
The idea we are exploring here is one of compounding distributions. A compound probability distribution $H$ results when a parametrized distribution $F$ is marginalized given a distribution for the parameter. In our case, we have a single parameter $L$ (univariate), and thus:
$$f_D(t) = \int f_N(t \mid \ell) \; f_L(\ell) \, d\ell,$$
where $f_N(t \mid \ell)$ is normal with mean $\mu_D \ell$ and variance $\sigma_D^2 \ell$ and $f_L(\ell)$ is normal with mean $\mu_L$ and variance $\sigma^2_L$.
I don't know much about compounding distributions. From what I've read, compounding a normal distribution with a normally-distributed mean leads to a another normal distribution. In our case, we have a normally-distributed mean and normally-distributed variance and they depend on the same underlying normal parameter $\ell$. It would be exciting to find out that such a compounding led to a normal distribution.