Let $B(p)$ denote the Bernoulli distribution over $\{0,1\}$ and $B(p)^n$ the corresponding product distribution over $\{0,1\}^n$. For $n>1$ and $0<x<1$, define $$P_n(x):=B(\frac12+\frac x2)^n$$ and $$Q_n(x):=B(\frac12-\frac x2)^n.$$ I can show $ ||P_n(x)-Q_n(x)||_1 \le 2.1\sqrt{n}x$ using Pinsker's inequality.
Conjecture: $$ ||P_n(x)-Q_n(x)||_1 \le \sqrt{3n}x.$$
Looking at [Pinsker's inequality], I am assuming that the left-hand side of the conjectured inequality is understood as the the total variation norm of the signed measure $P_n(x) - Q_n(x)$, which equals \begin{equation} s(y):=s_n(y):=2\sum_{j=0}^m\binom nj\big[\big(\tfrac12 + y\big)^{n - j} \big(\tfrac12 - y\big)^j - \big(\tfrac12 - y\big)^{n - j} \big(\tfrac12 + y\big)^j\big], \end{equation} where \begin{equation} m:=\lfloor n/2\rfloor,\quad y:=x/2\in[0,1/2). \end{equation} Using identities $(n-j)\binom nj=n\binom{n-1}j$ and $j\binom nj=n\binom{n-1}{j-1}$, we have \begin{equation} \tfrac1{2n}\,s'(y):=\sum_{j=0}^m\binom{n-1}j\big[\big(\tfrac12 + y\big)^{n - j-1} \big(\tfrac12 - y\big)^j + \big(\tfrac12 - y\big)^{n - j-1} \big(\tfrac12 + y\big)^j\big] \end{equation} \begin{equation} -\sum_{j=1}^m\binom{n-1}{j-1}[\big(\tfrac12 + y\big)^{n - j} \big(\tfrac12 - y\big)^{j-1} + \big(\tfrac12 - y\big)^{n - j} \big(\tfrac12 + y\big)^{j-1}\big]. \end{equation} Making in the second sum the substitution $j=k+1$ and then replacing there $k$ back by $j$, we have \begin{equation} \tfrac1{2n}\,s'(y)\Big/\binom{n-1}m=\big(\tfrac12 + y\big)^{n - m-1} \big(\tfrac12 - y\big)^m + \big(\tfrac12 - y\big)^{n - m-1} \big(\tfrac12 + y\big)^m =2^{2-n}(1-4y^2)^{n-m-1}, \end{equation} which is non-increasing in $y\in[0,1/2)$ for $n=1,2,\dots$. So, $s$ is concave.
Moreover, \begin{equation} s'_n(0)=n2^{3-n}\binom{n-1}{\lfloor n/2\rfloor}\le 2A\sqrt n \tag{*} \end{equation} for some universal real constant $A>0$. It should be easy to see that one can take ($A=2$ for $n=1$ and) $A=\sqrt3$ for $n=2,3,\dots$.
Since $s(0)=0$ and $s$ is concave, it will then follow that for $n=2,3,\dots$ $$\|P_n(x) - Q_n(x)\|=s(y)\le2Ay\sqrt n=\sqrt{3n}\,x,$$ as desired (actually, for all $x\in[0,1]$).
Addendum: To verify $(*)$, let \begin{equation} r_n:=s'_n(0)/(2\sqrt n). \end{equation} We need to show that $r_n\le\sqrt3$ for $n=2,3,\dots$. For $m=1,2,\dots$, \begin{equation} \frac{r_{2m+2}}{r_{2m}}=\sqrt{\frac{4m^2+4m+1}{4m^2+4m}}>1, \end{equation} so that $r_{2m}$ increases in $m$, and $r_{2m}\to2\sqrt{2/\pi}$ as $m\to\infty$. So, $r_{2m}<2\sqrt{2/\pi}<\sqrt3$ for all $m=1,2,\dots$. Similarly, in the "odd" case, for $m=2,3,\dots$ \begin{equation} \frac{r_{2m+3}}{r_{2m+1}}=\sqrt{1-\frac{6 m^3+11 m^2-8 m-4}{4(2m+1)(m^2-1)^2}}<1, \end{equation} so that $r_{2m+1}$ decreases in $m\ge2$. It remains to note that $(r_1,r_3,r_5)=(\sqrt4, \sqrt3, \sqrt{45/16})$, so that $r_1>r_3>r_5$.
