One of the postulates of Quantum Mechanics involves the so-called Born's Rule:
Formulated by Max Born in 1926, it gives the probability that a measurement of a quantum system will yield a given result. In its simplest form, it states that the probability density of finding a particle at a given point, when measured, is proportional to the square of the magnitude of the particle's wave function at that point.
The wave function is a rapidly-decreasing function, $\psi(\mathbf{x}, t) \in \mathscr{S}\left(\mathbb {R} ^{3},\mathbb {C} \right) \times \mathbb{R}$, where
$$ \mathscr{S}\left(\mathbb {R} ^{n},\mathbb {C} \right):=\left\{f\in C^{\infty }(\mathbb {R} ^{n},\mathbb {C} )\mid \forall \alpha ,\beta \in \mathbb {N} ^{n},\|f\|_{\alpha ,\beta }:= \sup _{x\in \mathbb {R} ^{n}}\left|x^{\alpha }(D^{\beta }f)(x)\right|<\infty \right\}$$
is the Schwartz space.
Problem. $\mathscr{S}$ is a Fréchet space! Therefore, it's not possible to define a "norm" ($\|f\|_{\alpha, \beta}$ is just a seminorm).
And yet, it's needed in order $\|\psi(\mathbf{x},t)\|^2 d\mathbf{x}$ to be seen as a probability density.
Remark. $\mathscr{S}(\mathbb{R}^n, \mathbb{C})\hookrightarrow L^2(\mathbb{R}^n, \mathbb{C})$, but I don't think this is of any help.
Best Answer
$L^2$ is the completion of $\mathscr S$ in $||\cdot||_2$:
Whenever you study wave-functions in Quantum Mechanics, you start with some function space $\mathcal F(\mathbb R^n, \mathbb C) \times \mathbb R$ and the $L^2$-norm $$||\psi(t)||^2_2 = \int d\mathbf x~ |\psi(\mathbf x, t)|^2 $$ There are different function spaces for different situations, for example $\mathcal C_c^2(\mathbb R^n, \mathbb C)$ (twice differential fcts. with compact support), $\mathcal C_c^\infty(\mathbb R^n, \mathbb C)$ (smooth fcts. with compact support) or $\mathscr S(\mathbb R^n, \mathbb C)$ (as you defined it). All of these spaces are dense subsets of $L^2$, meaning they contain enough functions to approximate any $L^2$-function to arbitrary precision.
However, Hilbert spaces are complete space. This means:
The Hilbert space $\mathscr H(\mathbb R^n, \mathbb C)$ of QM is the completion of the function space $\mathcal F(\mathbb R^n, \mathbb C)$!
But since $\mathcal F(\mathbb R^n, \mathbb C) \subset L^2(\mathbb R^n, \mathbb C)$ is dense, the only completion in the $L^2$-norm is $L^2$ itself, i.e.: $$\mathscr H(\mathbb R^n, \mathbb C) := \overline{\mathcal F(\mathbb R^n, \mathbb C)} = L^2(\mathbb R^n, \mathbb C). $$
Why don't we start with $L^2(\mathbb R^n, \mathbb C)$ ?
$L^2$ may be the largest space of functions with finite $L^2$-norm, but it consequently has the weakest properties. Simple quantum mechanical operators are not necessarily defined on it. For example, consider $\psi \in L^2(\mathbb R)$. Then, its $\hat x$ and $\hat p$ expectation values might not be defined: $$ \langle \hat x \rangle_\psi = \langle \psi(x), \hat x \psi(x) \rangle = \int dx~ x |\psi(x)|^2 = \infty,~ \langle \hat p\rangle_\psi = \int dx~ \overline{\psi(x)} \psi'(x) = \infty. $$ It could be that $\psi'(x)$ is not defined in the first place. This is why one starts with a safer space (e.g. $\mathscr{S}$) and then considers the completion of the operator ($\hat x$ or $\hat p$ or some other) on this smaller space.
I hope that clarifies things!