When $n=1$, obviously $||x||_p = |x_1|$ which is the limit in question. Now assume that $n>1$ and $x = (x_1,...,x_n)\in \mathbb{R}^n$ has at least one non-zero coordinate (otherwise the limit is $0$).
In case $x$ has only one non-zero coordinate, the problem reduces to the case of $n=1$, and the limit becomes equal to the absolute value of that non-zero coordinate. Hence we will assume that $x$ has at least two non-zero coordinates, and will prove that in this case the limit is infinite. Indeed, assume without loss of generality that $|x_1| \geq |x_i|$ for all $i=2,...,n$ and that $x_2 \neq 0$.
Then
$$
||x||_p = |x_1| \left( 1 + \frac{|x_2|^p}{|x_1|^p} +...+ \frac{|x_n|^p}{|x_1|^p} \right)^{1/p} \geq |x_1| \left( 1 + \frac{|x_2|^p}{|x_1|^p} \right)^{1/p} := |x_1| \left( 1 + t^p \right)^{1/p}, \tag{1}
$$
where $t = |x_2|/|x_1| $ and by our assumption $0<t \leq 1$.
We claim that $(1 + t^p)^{1/p} \to \infty$ as $p \to 0$, which clearly implies by $(1)$ that $||x||_p \to \infty$ as $p\to 0$. To prove the claim we use the inequality
$$
\log(1 + t ) \geq t - \frac{t^2}{2}, \text{ for all } 0 \leq t \leq 1. \tag{2}
$$
To see $(2)$, consider the function $f(t) = \log(1 + t) - t + t^2/2$ in $[0,1]$ and differentiate twice to see that it increases.
Now, using $(2)$ we get
$$
\log( 1 + t^p)^{1/p} \geq \frac{1}{p} \left( t^p - \frac{t^{2p}}{2} \right) = \frac{t^p}{p} \left( 1 - \frac{t^p}{2} \right) \to \infty \text{ as } p\to 0.
$$
This completes the proof that $||x||_p \to \infty$ when $p\to 0$, as long as $x$ has at least two non-zero coordinates.