We shall prove the inequality 
\begin{equation*}
	\sum_{i=1}^n|a_i|^2+\sum_{i=1}^n|b_i|^2\ge\frac Cn \sum_{i,j=1}^n|a_ib_j| \tag{0}
\end{equation*}
with $C:=4/\sqrt3=2.309\dots$. 
We use the notations $|a|:=\|a\|_2$ and $ab:=a^Tb$. Without loss of generality, the $a_i$ and $b_j$'s are nonzero vectors. 

For two nonzero vectors $a$ and $b$, let $d(a,b)\in[0,\pi/2]$ denote the angle between the straight lines carrying the vectors $a$ and $b$. The function $d$ is a metric, since the big circles are the geodesic lines on the 2D sphere. 

For $i,j$ in $[n]:=\{1,\dots,n\}$, let then
$$d_{ij}:=d(a_i,b_j)=\arccos c_{ij},\quad c_{ij}:=\frac{|a_ib_j|}{|a_i|\,|b_j|},$$
so that $d_{ij}\in[0,\pi/2]$ is the angle between the straight lines carrying the vectors $a_i$ and $b_j$. 

Take any $i,j,k$ in $[n]:=\{1,\dots,n\}$. Since $a_ib_i=0$ and $d$ is a metric, 
\begin{equation*}
	|d_{ki}-\pi/2|=|d_{ki}-d_{ii}|\le d(a_i,a_k)=:t
\end{equation*}
and hence 
\begin{equation*}
	|a_kb_i|\le|a_k|\,|b_i|\sin t. \tag{1}
\end{equation*}
Moreover, again because $d$ is a metric, 
\begin{equation*}
	t\le d_{ij}+d_{kj}. \tag{2}
\end{equation*}

If $d_{ij}+d_{kj}\ge\pi/2$, then $d_{kj}\in[\pi/2-d_{ij},\pi/2]\subseteq[0,\pi/2]$ and hence 
$c_{ij}^2+c_{kj}^2\le\cos^2 d_{ij}+\cos^2(\pi/2-d_{ij})=1\le5/4$, so that 
\begin{equation*}
	c_{ki}^2+c_{ij}^2+c_{kj}^2\le9/4. \tag{3}
\end{equation*}
If $d_{ij}+d_{kj}<\pi/2$, then (2) implies $\sin t\le\sin(d_{ij}+d_{kj})$. So, by (1), 
\begin{equation*}
	c_{ki}\le c_{kj}\sqrt{1-c_{ij}^2}+c_{ij}\sqrt{1-c_{kj}^2}.
\end{equation*}
Now the Cauchy--Schwarz inequality yields
\begin{equation*}
	c_{ki}^2\le(c_{kj}^2+c_{ij}^2)(2-c_{kj}^2-c_{ij}^2).
\end{equation*}
The latter inequality together with the conditions that $c_{ki}^2,c_{kj}^2,c_{ij}^2$ are in $[0,1]$ implies (3). Thus, (3) holds for any $i,j,k$. 

Therefore,
\begin{equation*}
	\frac94\,n^3\ge\sum_{i,j,k\in[n]}(c_{ki}^2+c_{ij}^2+c_{kj}^2)
	=3n\sum_{i,j\in[n]}c_{ij}^2,
\end{equation*}
so that 
\begin{equation*}
	\sum_{i,j\in[n]}c_{ij}^2\le\frac34\,n^2,
\end{equation*}
which further implies 
\begin{align*}
	\sum_{i,j\in[n]}|a_ib_j|&=\sum_{i,j\in[n]}c_{ij}|a_i|\,|b_j| \\ 
	&\le\sqrt{\sum_{i,j\in[n]}c_{ij}^2}
	\sqrt{\sum_{i,j\in[n]}|a_i|^2\,|b_j|^2} \\ 
	&=\sqrt{\sum_{i,j\in[n]}c_{ij}^2}
	\sqrt{\sum_{i\in[n]}|a_i|^2}\,\sqrt{\sum_{j\in[n]}|b_j|^2} \\ 
	&\le\sqrt{\frac34\,n^2}\times\frac12\,\Big(\sum_{i\in[n]}|a_i|^2+\sum_{j\in[n]}|b_j|^2\Big),
\end{align*}
so that we do have (0) with $C=4/\sqrt3$.