It is true in general. A trivial direction is if $\mathfrak p$ splits in $L/K$, then it splits in any sub-extension because the spliting means unramified and no residue extensions. In particular $\mathfrak p$ splits in all $L_i$'s.
Suppose now that $\mathfrak p$ splits in all $L_i$. Without lose of generalities, we can suppose $n=2$. As the problem is local at $\mathfrak p$, we can localize and suppose $K$ is a discrete valuation field (you can stick to number fields if you prefer). Denote by $O_K$ the valuation ring of $K$ and by $O_{L_i}$ the integral closure of $O_K$ in $L_i$. Let $\pi$ be a uniformizing element of $O_K$. By hypothesis $O_{L_i}/(\pi)$ is a direct sum of copies of $k$, the residue field of $K$.
Consider the tensor product $A=O_{L_1}\otimes O_{L_2}$ over $O_K$. Its generic fiber $A\otimes K$ is $L_1\otimes_K L_2$ and is reduced because $L_i/K$ is separable. And $A/\pi A=O_{L_1}/(\pi) \otimes_k O_{L_2}/(\pi)$ is a direct sum of copies of $k$. I claim that $O_L$ is a quotient of $A$. This will imply that $O_L/(\pi)$ is a direct sum of copies of $k$ hence $\mathfrak p$ splits in $L$.
Proof of the claim. Consider the canonical map $f : A\otimes K=L_1\otimes L_2\to L, \quad x_1\otimes x_2\mapsto x_1x_2.$ The image $f(A)$ is a subring of $L$, finite over $O_K$ because $A$ is finite over $O_K$. Hence $f(A)\subseteq O_L$. Moreover $f(A)/(\pi)$ is a quotient of $A/\pi A$, so it is a direct sum of copies of $k$. In particular $O_K$ is unramified in $f(A)$, so $f(A)$ is regular hence equal to $O_L$.