What is the best you can do with a noisy message?
Suppose that $k \ll n$. In that case, your friend can send you the index of the first door containing a treasure. Of the $k$ numbers you get, you pick the smallest one. If $k$ is small enough compared to $n$, then it is very likely that the smallest of the $k$ numbers is the one that your friend sent you.

More concretely, suppose that $k$ is constant. With probability $1-1/n$, the number that your friend sends you is at most $\log n$. The probability that the other $k-1$ numbers are more than $\log n$ is $(1-\log n/n)^{k-1} \approx 1-k\log n/n$. We conclude that this strategy fails with probability at most roughly $1/n + k\log n/n$, which tends to $0$ as $n \to \infty$.

To determine how small $k$ should be, let us be a bit more careful. Denote by $m$ the number that your friend sends you; the distribution of $m$ is geometric (we ignore for now the event in which there are no treasures at all), and so $\Pr[m] = 2^{-m}$. The probability that one of the other $k-1$ numbers is smaller than $m$ is $1-(1-(m-1)/n)^{k-1}$. Overall, this strategy fails with probability
$$
\frac{1}{2^n} + \sum_{m=1}^n 2^{-m} \left(1 - \left(1 - \frac{m-1}{n}\right)^{k-1}\right) = 1 - \sum_{m=1}^n 2^{-m} \left(1 - \frac{m-1}{n}\right)^{k-1}.
$$
We can non-rigorously estimate this as
$$
1 - \sum_{m=1}^\infty 2^{-m}e^{-(m-1)k/n} = 1 - \frac{1}{2-e^{-k/n}} = \frac{1-e^{-k/n}}{2-e^{-k/n}} \approx \frac{k}{n}.
$$
This suggests that if $k = o(n)$ then you win with probability $1-o(1)$.