The answer is no. Let $B=(b_{ij}\colon i\in J_n,\;j\in J_n)$, where 
\begin{equation}
	b_{kj}:=\frac1n\sqrt{\frac2n}\cos\frac{2\pi kj}n
\end{equation}
for $k$ and $j$ in $J_n:=\{0,\dots,n-1\}$; it will be more convenient for us to use the index set $J_n$ instead of $[n]:=\{1,\dots,n\}$. 
Using Euler's formula $\cos t=\frac12\,(e^{it}+e^{-it})$, it is straightforward to check that the matrix $nB$ is orthogonal. Here are these calculations in Mathematica: 

[![enter image description here][1]][1]


Hence, 
\begin{equation}
	\Big|\sum_{i,j=0}^{n-1}b_{ij}x_iy_j\Big|=|x^T By|\le\|x\|_2\|By\|_2
	=\frac1n\,\|x\|_2\|y\|_2=1
\end{equation}
if $|x_i|=|y_j|=1$ for all $i,j\in J_n$. So, $\|B\|_2\le1$. 
Let $X=(x_{ij}\colon i\in J_n,\;j\in J_n)$ with $x_{ij}=1(i=j)$, so that $\|X\|_1=1$. 

However, 
\begin{equation}
	\sum_{i,k\in J_n}^n\Big|\sum_{j\in J_n}x_{kj}b_{ij}\Big|
	=\sum_{i,k\in J_n}|b_{ik}|
	=\frac1n\sqrt{\frac2n}\sum_{j,k\in J_n}\Big|\cos\frac{2\pi kj}n\Big|\ge c\sqrt n \tag{1}
\end{equation}
for some universal real constant $c>0$ and all natural $n$. So, your $C(n)$ goes to $\infty$. 

---

**Details on the inequality in (1):** Let $P_n$ denote the set of all pairs $(k,j)\in J_n^2$ such that the fractional part of $\dfrac{kj}n$ is in the interval $[0,1/6]$. Take any natural $n\ge100$, any natural $k\in[n/24,n/12]$, and any integer $m\in[0,(n-5)/24]$. Then the interval $\Big[\dfrac{mn}k,\dfrac{(m+1/6)n}k\Big]$ is of length $n/(6k)\ge2$ and contained in the interval $[0,n-1]$. So, there is some $j_{k,m}\in J_n\cap\Big[\dfrac{mn}k,\dfrac{(m+1/6)n}k\Big]$, and then 
\begin{equation}
\dfrac{kj_{k,m}}n\in[m,m+1/6], \tag{2}	
\end{equation}
so that $(k,j_{k,m})\in P_n$. Moreover, by (2), for any given natural $n$ and $k$, the values of $j_{k,m}$ are distinct for distinct values of $m$. 
Each of the intervals $[n/24,n/12]$ (for the values of $k$) and $[0,(n-5)/24]$ (for the values of $m$) contains $\sim n/24$ integers (as $n\to\infty$). It follows that the cardinality $|P_n|$ of the set $P_n$ is $\gtrsim(n/24)^2$. So,
\begin{equation}
\sum_{j,k\in J_n}\Big|\cos\frac{2\pi kj}n\Big|
\ge\sum_{(k,j)\in P_n}\cos\frac{2\pi kj}n\gtrsim(n/24)^2/2, 
\end{equation}
since $\cos\dfrac{2\pi kj}n\ge1/2$ for all $(k,j)\in P_n$. 

Thus, the inequality in (1) follows. 


  [1]: https://i.sstatic.net/rym7T.png