I believe we have to distinguish the finite-dimensional from the infinite dimensional case.
Regardless, if $0<p<1$, $\|x\|_p := (\sum |x_i|^p)^{\frac 1 p}$ is not a norm as it fails to satisfy the triangle inequality. That's why we use instead the metric $d(0,x) = \sum |x_i|^p = \|x\|_p^p$ to define the topology and remark that this is also not a norm since it is homogeneous of degree $p$.
I do not know how the balls defined by this metric look like, but they ought to be convex because the metric satisfies the triangle inequality, right? In other words, they can't be the corresponding astroid-shaped superellipses for $0<p<1$. So why is $\ell^p$ said to not be locally convex (at least in the infinite dimensional case)?
Here's a proof why $l^p(\mathbb N)$ is not locally convex, this is just for simplicity, it can be easily generalized.
If it would be locally convex, then the unit ball $B_1(0)$ would contain a convex neighborhood U of $0$. Then there must be $\delta>0$ with $B_{2\delta}(0)\subset U$, hence also $\mathrm{conv}(B_{2\delta}(0))\subset U\subset B_1(0)$.
Let $e_i := (\underbrace{0,...,0,1}_{i},0,0,...)$. Then $\delta^{\frac 1p} e_i\in B_{2\delta}(0)$. We get $\sum_{i=1}^n \frac{1}{n}\delta^{\frac 1p} e_i\in \mathrm{conv}(B_{2\delta}(0))\subset B_1(0)$ for all $n\in\mathbb N$, which means $1>\|\sum_{i=1}^n \frac{1}{n}\delta^{\frac 1p} e_i\|_p^p = \delta n^{1-p}$ for all $n\in\mathbb N$ which is a contradiction because $1-p> 0$.