Metric Spaces: Why $L_\infty$ selects the maximum value
In the general case, let  $x = (x_1,\dots,x_n)$ be a finite-length vector (in a finite dimensional space). 
The finite sequence of absolute values $|x_i|$ does attain its maximum (because the sequence is finite), denoted $M = \max_i |x_i|$. 

Let $m$ be the (exact) number of coordinates in $x = (x_1,\dots,x_n)$ whose absolute value  is equal to $M$. Thus, $1\le m\le n$. Then, we can lower and upper bound the $\ell_p$ norm of $x$ as follows: $$(mM^p)^{\frac{1}{p}}\le\ell_p(x)\le (nM^p)^{\frac{1}{p}}\,.$$

Both $m^\frac{1}{p}$ and $n^\frac{1}{p}$ tend to $1$ as $p\to\infty$, thus $\ell_p(x) \to M$, by the squeeze theorem.

[EDIT] To provide more concrete substance, let us see what happen with your example: $x=[1,2,3]^T$ (the transposition does not change the result): $$\|x\|_p = d_p(x,0) = (1^p+2^p+3^p)^{1/p} = 3\times\left(\left(\frac{1}{3}\right)^p+\left(\frac{2}{3}\right)^p+1\right)^{1/p}\,.$$

Both $\left(\frac{1}{3}\right)^p$ and $\left(\frac{2}{3}\right)^p$ tend to $0$ as $p \to \infty$. Thus, $\|x\|_p \to 3$.