Yes, it can. You can ask more generally for the probability of $k$ successes for any $k$, so we're looking at a sequence of binomial distributions $X_n \sim B(n, \frac{1}{n})$. We have
$$\mathbb{P}(X_n = k) = {n \choose k} \frac{1}{n^k} \left( 1 - \frac{1}{n} \right)^{n-k}$$
and as $n \to \infty$ this converges to $e^{-1} \frac{1}{k!}$. These limits themselves are non-negative reals adding up to $1$ so also describe some distribution, namely the Poisson distribution $X \sim \text{Pois}(1)$ with parameter $1$; our calculations are saying that the sequence of binomial distributions $X_n$ converges to $X$ (more or less). This is an important special case of the Poisson limit theorem, and is "where the Poisson distribution comes from." In particular $e^{-1}$ is the probability $\mathbb{P}(X = 0)$.
The sample space is just $\mathbb{N}$. Of course when we pass to the binomial distribution we flatten the picture of the situation; we pass from a bunch of uniform distributions to Bernoulli distributions to their sum. Your comment about the sample space looking like $\mathbb{N}^{\mathbb{N}}$ in the limit can be made precise as follows: the process of taking $n$ random samples from the uniform distribution on $\{ 1, 2, \dots n \}$ itself also has a limit in a suitable sense, namely a (discrete) Poisson point process. The number of samples contained in any subset
$$S \subseteq \{ 1, 2, \dots n \}$$
of size $s$ is a binomial distribution $B(n, \frac{s}{n})$, and as $n \to \infty$ (keeping $s$ fixed) this has a limit $\text{Pois}(s)$. So we can think about the limit process as a process of "choosing infinitely many random points from $\mathbb{N}$," in such a way that the distribution of the number of points in any subset $S \subset \mathbb{N}$ of size $s$ is $\text{Pois}(s)$. This works out to be equivalent to saying that the number of points equal to any $n \in \mathbb{N}$ is $\text{Pois}(1)$, and each of these Poisson distributions is independent. Surprisingly we can make sense of this even though there is no uniform distribution on $\mathbb{N}$! (Note however that we are only counting points, so for us the points have become indistinguishable; there is no "first point," "second point," etc. so no marginal distribution over the locations of such points, which would have to be the nonexistent uniform distribution on $\mathbb{N}$.)
A continuous variant of this limiting construction is to consider taking $n$ random samples from the uniform distribution on $[0, n]$. We can recover the finite sample spaces above by rounding up. Here the analogue of the computation above is that the number of samples in any interval of length $\lambda$ is $B(n, \frac{\lambda}{n})$, and as $n \to \infty$ this converges to $\text{Pois}(\lambda)$. The limit is a Poisson point process on $[0, \infty)$, "choosing infinitely many random points from $[0, \infty)$," where the number of points in any interval of length $\lambda$ is $\text{Pois}(\lambda)$. This is used to model, say, rain, where the random points correspond to raindrops.
