$\newcommand{\mk}{\mathfrak}$ Let $\mk{h}_{2n+1}$ be the $(2n+1)$-dimensional Lie algebra (basis $x_1,\dots,x_n,y_1,\dots,y_n,z$, nonzero brackets $[x_i,y_i]=z$).
The $n$-dimensional abelian Lie algebra $\mk{a}_n$ (basis $a_1,\dots,a_n$) acts on it by $a_i\cdot x_i=x_i$, $a_i\cdot y_i=-y_i$, rest of the action being zero.
Through the quotient map $\mk{h}_{2n+1}\to\mk{a}_{2n}$ one thus get an action of $\mk{h}_{2n+1}$ on $\mk{h}_{4n+1}$.
Then define the semidirect product $\mk{h}_{2n+1}\ltimes\mk{h}_{4n+1}$, and identify the basis central element of the two.
This thus defines a $(6n+1)$-dimensional Lie algebra with basis $x_i,y_i,t_i,u_i,v_i,w_i,z$ ($1\le i\le n$) and nonzero brackets $$[x_i,t_i]=t_i,\quad [x_i,v_i]=-v_i,\quad [y_i,u_i]=u_i, \quad[y_i,w_i]=-w_i,$$ $$[x_i,y_i]=[t_i,v_i]=[u_i,w_i]=z.$$
This is supersolvable. The nilpotent radical has basis $t_i,u_i,v_i,w_i,z$ ($1\le i\le n$), so the quotient is abelian with basis $x_i,y_i$ ($1\le i\le n$). Observe that the isomorphic image of $\mk{h}_{2n+1}$, i.e., with basis $x_i,y_i,z$ ($1\le i\le n$) is a Cartan subalgebra (= nilpotent, equal to its normalizer).
This does not split (for $n\ge 1$). Indeed, suppose by contradiction it does: let $\mk{a}$ be a splitting subalgebra. Then the subspace generated by $\mk{a}$ and $z$ is an abelian $(2n+1)$-dimensional subalgebra and is a Cartan subalgebra as well, unlike the previous one which is abelian. But in a solvable Lie algebra all Cartan subalgebras are conjugate (cf Bourbaki) under automorphisms of the Lie algebra. This is a contradiction (for $n=1$ it can be directly checked: indeed $[x_1+a,y_1+b]$ is never zero for any $a,b$ in the nilpotent radical).