How to generalize this version of Tarski’s Fixed Point Theorem?
The "baby version" is not sufficient to obtain your result for $n > 1$ as a corollary. This is possible only in case that $f : [0,1]^n \to [0,1]^n$ has the special form $$f(x_1,\dots,x_n) = (\phi_1(x_1),\dots,\phi_n(x_n))$$ with continuous increasing $\phi_i : [0,1] \to [0,1]$. But in general $f$ has the form $$f(x_1,\dots,x_n) = (f_1(x_1,\dots,x_n),\dots,f_n(x_1,\dots,x_n))$$ where each $f_i :[0,1]^n \to [0,1]$ depends on all variables.
In fact, your result is true, but you will need a new proof. See Asaf Karagila's answer.
Let me close with a remark concerning your strategy to find a fixed point. It is not expedient to pick some $x_0 =(x^0_1,\dots,x^0_n)$ and to consider the functions obatined by fixing all but one coordinate. Let us see what happens for $n=2$. You consider the function $f^1 : [0,1] \to [0,1], f^1(x) = f_1(x,x^0_2)$ and conclude that it has a fixed point $\xi_1$, i.e. $f_1(\xi_1,x^0_2) = \xi_1$. Next you consider $f^2 : [0,1] \to [0,1], f^2(x) = f_1(x^0_1,x)$ and conclude that it has a fixed point $\xi_2$, i.e. $f_2(x^0_1,\xi_2) = \xi_2$. But there is no reason why you should have $f(\xi_1,\xi_2) = (\xi_1,\xi_2)$.
The Knaster–Tarski fixed point theorem states that:
Suppose that $(L,\leq)$ is a complete lattice and $f\colon L\to L$ is increasing, i.e. $x\leq y\implies f(x)\subseteq f(y)$. Then $\{x\in L\mid f(x)=x\}$ is a complete lattice under the induced order.
Note that a complete sublattice cannot be empty, since $\varnothing$ must have a supremum and infimum (which make the bottom and top elements of the lattice).
Note that $[0,1]$ is a complete lattice, and $[0,1]^n$ is also a complete lattice. Granted, now, most uses of this theorem really just care that there is a fixed point. This is commonly used in the proof of the Cantor–Bernstein theorem when we want to find a set closed under some function.
Just finding a fixed point is easy. Let $D=\{x\in L\mid x\geq f(x)\}$ and let $s=\inf D$. Even if $D$ is empty, it still has a infimum, it would just be the top element of $L$. This is because $L$ is a complete lattice.
Now, if $s<x$, then $f(s)\leq f(x)$. Since this holds for all $x\in D$, it follows that $f(s)\leq s$, by the virtue of being an infimum. Therefore $s\in D$, and it is in fact a minimum. But now we have that $f^2(s)\leq f(s)\leq s$. So $f(s)\in D$ as well, which means that $s\leq f(s)$—again by the virtue of being an infimum—and therefore $f(s)=s$. We can even show that $s$ is the smallest fixed point, i.e. the bottom element of the fixed points lattice.
If we assume continuity of $f$, in the sense that it respects the $\sup$ operator, we get an even easier proof of what is known as the Kleene fixed points theorem.
Let's use $0$ to denote the bottom element of $L$. Then $0\leq f(0)\leq f^2(0)\leq ...$, let $x=\sup\{f^n(0)\mid n\in\Bbb N\}$, which exists since $L$ is a complete lattice. Since $f$ is continuous we get that $$f(x)=f(\sup\{f^n(0)\mid n\in\Bbb N\}) = \sup\{f(f^n(0)\mid n\in\Bbb N\}=\sup\{f^{n+1}(0)\mid n\in\Bbb N\}=x.$$
Finally, we apply these proofs to your question. And you can see that this is essentially the same proof for both $[0,1]$ and $[0,1]^n$. And in fact, much much more.