Calling $\langle X_1,X_2,X_3\rangle$ a random permutation we implicitly assume the existence of an underlying probability space $(\Omega,\mathcal A,\mathsf{P})$.
The $X_i$ are looked at as functions $\Omega\to\mathbb R$, and the probability space can be constructed in such a way that function $\langle X_1,X_2,X_3\rangle:\Omega\to\mathbb R^3$ only takes values in the set: $$S=\{\langle x_1,x_2,x_3\rangle\mid\{x_1,x_2,x_3\}=\{1,2,3\}\}$$
Then element $\langle x_1,x_2,x_3\rangle\in S$ induces the event: $\{\langle X_1,X_2,X_3\rangle=\langle x_1,x_2,x_3\rangle\}$ which is actually the preimage of $\{\langle x_1,x_2,x_3\rangle\}$ under function $\langle X_1,X_2,X_3\rangle:\Omega\to\mathbb R^3$: $$\langle X_1,X_2,X_3\rangle^{-1}(\{\langle x_1,x_2,x_3\rangle\})=\{\omega\in\Omega\mid \langle X_1(\omega),X_2(\omega),X_3(\omega)\rangle=\langle x_1,x_2,x_3\rangle\}$$
In this situation we can choose for $\Omega=S$ and we can prescribe $X_i:\Omega=S\to\mathbb R$ by: $$\langle x_1,x_2,x_3\rangle\mapsto x_i$$
Then we can take $\mathcal A=\wp(S)$ and - if we want equiprobability for all outcomes - we can determine $\mathsf{P}$ by stating that $\mathsf{P}(\{s\})=\frac16$ for every $s\in S$ (note that $S$ has $6$ elements).
I hope that this gives some clarity. My main point is that the $X_i$ are functions and the $x_i$ are not. Quite often in situations like this not much attention goes to the underlying probability space, and in fact it is a good thing that that is not really needed. It is enough to know that we can model the situation, but how that could be done is irrelevant in most situations. A nice question on that subject you can find here.