It's true: there's an isometric embedding of every real tree $T$ on some $\ell^1$-space, i.e., a set $A$ and a map $f:T\to \ell^1(A)$ that is an isometric embedding, that is, satisfies $\|f(x)-f(x')\|=d(x,x')$ for all $x\in T$.
I don't know a reference; here's a proof. Let $(x_t)_{t<\alpha}$ be an enumeration of points on the real tree $T$ by an ordinal $\alpha$, assume that the convex hull of them is $T$.
Let $T_\beta$ be the closed convex hull of $\{x_t:t<\beta\}$ (so $T_\alpha=T$). Let us construct by induction an isometric embedding $f_\beta:T_\beta\to\ell^1(\beta)\subset\ell^1(\alpha)$, so that whenever $\beta\le\gamma$ then $f_\gamma$ extends $f_\beta$.
For $\beta=1$, $T_\beta=\{x_0\}$, and we just prescribe $f_\beta(x_0)=0$. For $\beta=\gamma+1>1$ a successor ordinal, $T_\beta$ is the convex hull of $T_\gamma\cup\{x_\gamma\}$. Let $y_\gamma$ be the projection of $x_\gamma$ on $T_\gamma$. So $$T_\beta=T_\gamma\cup [y_\gamma,x_\gamma].$$ We define $f_\beta$ so as to extend $f_\gamma$ on $T_\gamma$, and prescribe $f_\beta(y)=f_\gamma(y_\gamma)+d(y,y_\gamma)e_\beta$ for every $y\in [y_\gamma,x_\gamma]$, where $(e_t)$ is the canonical basis of $\ell^1(\alpha)$ and $d$ is the distance in the tree. This is well-defined (note that $f_\gamma(y_\gamma)$ is defined twice) and defines an isometric embedding. Finally, for $\beta$ a limit ordinal, then $T_\beta$ is the closure of $T'_\beta=\bigcup_{\gamma<\beta}T_\gamma$, we just define $f_\beta$ on $T'_\beta$ by saying its graph is the union of graphs of all $f_\gamma$ for $\gamma<\beta$; this is an isometric embedding into a complete metric space, so extends uniquely to an isometric embedding of the closure $T_\beta$ into $\ell^1(\beta)$.
So $f_\alpha$ is the required isometric embedding.