Finding the asymptotic distribution Let
$$Y_n=e^{\frac{-1}{n}\sum_{i=1}^nX_i}$$
with $X_1,...,X_n$ being $n$ i.i.d. Poisson random variables.
In order to find the distribution of $Y_n$ as $n\to\infty$, I first calculate
$$P(Y_n\le y)=P(e^{\frac{-1}{n}\sum_{i=1}^nX_i}\le y)=P(\ln{y^{-n}}\le \sum_{i=1}^nX_i)$$
Since $\sum_{i=1}^nX_i$ is Poisson distributed, the cdf of $Y_n$ can be found as
$$P(Y_n\le y)=1-e^{-n\lambda}\sum_{i=0}^{\lfloor\ln{y^{-n}}\rfloor}\frac{(n\lambda)^i}{i!}$$
but how should I proceed from here?
Edit:
$$\sqrt{n}(\ln{Y_n^{-1}}-\lambda)\to N(0,\lambda)$$
 A: Since the OP responded in the suggestions in the comments, let's see the elementary approach here:
$$Y_n=e^{-\frac{1}{n}\sum_{i=1}^nX_i} \implies \ln Y_n = -\bar X_n  $$
The Possion distribution (with parameter $\lambda$) does not have zero mean, so we must center in order to obtain the CLT
$$\ln Y_n + \lambda = -\bar X_n + \lambda = -(\bar X-\lambda)$$
We know that $\sqrt {n}(\bar X-\lambda) \to_d N(0,\lambda)$ and the same holds for its negative (due to the symmetry of the normal distribution).So
$$\sqrt {n}(\ln Y_n + \lambda) \to_d Z \sim N(0,\lambda)$$
which is a "asymptotic distribution" of $Y_n$, in the sense that we have found a function of $Y_n$ that converges to the zero-mean normal distribution.  Then approximately, we have that for large (but finite) samples, moving things around,
$$Y_n \sim_{approx.} e^W,\;\;\; W = \frac 1{\sqrt{n}}Z -\lambda \implies W \sim N(-\lambda, \lambda/n)$$
which tells us that for large but finite $n$, $Y_n$ follows  a log-normal distribution with the above underlying parameters.
At the limit, $Y_n$ uncentered and unscaled converges to a constant of course.
