The stochastic process $$ S_t = \exp\left( W_t - \frac{1}{2} t \right) $$ is a martingale (for example this could be seen by noting that it solves the SDE $dS_t = S_t dB_t$, which has no drift).
But this does not make intuitively sense for me, if $t$ is large, then $W_t - \frac{1}{2} t$ looks like the straight line $-\frac{1}{2}t$ with little perturbations, so this expression is very likely a large negative number, and therefore the exponential of this should be near zero. If it would be a martingale, then it expectation for each point in future should be the starting value, which in this case is $S_0 = \exp(0) = 1$ almost surely, and not $0$.
I also made some pictures of this process, using $R$ and the following code
N=1000
d=rnorm(N)
W=cumsum(d)
c=W-0.5*(1:N)
par(mfrow=c(2,2))
plot(W,type="l")
plot(c,type="l")
plot(exp(W),type="l")
plot(exp(c),type="1")
Which gives the output (the process $S_t$ is the one on the bottom on the right, and above is the process $W_t - t/2$, which resembles the straight line for large $t$):
So I do not understand, if it should be a martingale then it should not approach zero for large $t$, instead it should stay around its starting value (for most samples of course, but this process will approach zero for most samples, so this is not just a random deviation)?
I believe you are confusing the mean and mode of your distribution, which is asymetric.
$S_t$ follows a lognormal distribution with parameters $\mu = -\frac{1}{2} t$ and $\sigma^2 = t$. So, $E [S_t] = e^{\mu + \frac{1}{2} \sigma^2} = e^{-\frac{1}{2}t + \frac{1}{2} t} = e^0 = 1$, but the mode of the distribution is $e^{\mu - \sigma^2} = e^{-\frac{3}{2}t}$.
So, as $t \rightarrow \infty$, the mean remains $1$ (it is a martingale), but the mode goes to $0$. Intuitively, this is necessary to compensate unbounded values on the up side with bounded values on the down side.