This is intermediate between an answer, and a long comment on Max Alekseyev's post.
As discussed in the comments below his answer, there is a flaw in his post, so that it gives a lower bound. But there's a straightforward way to fix the flaw, after which the expected length is a calculation.
My attempt to do said calculation (below) contains an error somewhere, because it fails to obey the $n^{1/4}$ lower bound. The intermediate results are too oscillatory for Mathematica to test numerically, but the error should be easily fixable once found.
Begin with a slight generalization. Start with a single node of length $s$, and build a random tree as follows:
- Select a leaf node $n$ uniformly at random. Call its weight $w$.
- Select $p\in[0,1]$ uniformly at random.
- Attach two children to $n$ of weight $pw$ and $(1-p)w$.
Your problem asks: if $s=1$, what is the largest weight of a leaf, when there are $n$ nodes? Call this random variable $X^0_n$; in the general case, that length scales to $sX^0_n$.
As Max Alekseyev noticed, there is a natural recurrence structure to this tree. For notational convenience, suppose $n+1$ nodes. The root node has two children, weighted $P$ and $1-P$; let there be $K$ and $n-K$ nodes beneath each (inclusive); then $K$ is chosen uniformly from $\{0,1,\dots,n\}$. Take two iid copies of $X^0$, called $X^1$ and $X^2$. Since leaf nodes must descend from the children of the root, $$(X_{n+1}^0|K,P)=\max(PX^1_K,(1-P)X^2_{n-K})$$
To proceed exactly, introduce a CDF-like function. Let $F_n(t)=t\mathbb{P}\left[{X_n\leq\frac{1}{t}}\right]$; then \begin{align*} (n+1)F_{n+1}(t)&=(n+1)t\mathbb{P}\left[{X_n^0\leq\frac{1}{t}}\right] \\ &=\sum_{k=0}^n{\int_0^1{t\mathbb{P}\left[{\max(PX^1_K,(1-P)X^2_{n-K})\leq\frac{1}{t}}\middle|{K=k,|P-p|\leq dp}\right]}} \\ &=\sum_{k=0}^n{\int_0^1{t\mathbb{P}\left[{pX^1_k\leq\frac{1}{t}}\wedge{(1-p)X^2_{n-k}\leq\frac{1}{t}}\right]\,dp}} \\ &=\sum_{k=0}^n{\int_0^1{F_k(pt)F_{n-k}((1-p)t)\,\frac{dp}{t}}} \end{align*}
The finite sum is a discrete convolution, and can be eliminated by passing to generating functions. Let $\mathcal{F}(t,z)=\sum_{n=0}^{\infty}{F_n(t)z^n}$. Then $$\mathcal{F}(pt,z)\mathcal{F}((1-p)t,z)=\sum_{n=0}^{\infty}{\sum_{k=0}^n{F_k(pt)F_{n-k}((1-p)t)z^n}}$$ Integrating in $p$ recovers the recurrence from above, which simplifies to \begin{gather*} \frac{\mathcal{F}(t,z)}{\partial z}=\int_0^1{\mathcal{F}(pt,z)\mathcal{F}((1-p)t,z)\,\frac{dp}{t}} \\ \mathcal{F}(t,0)=F_0(t)=\begin{cases}t&0<t\leq1\\0&\text{otherwise}\end{cases} \end{gather*} Note that $F_n(t)=0$ for $t<0$, so that we can extend the bounds of integration to $\mathbb{R}$ without effect.
The other integral can be eliminated via the Fourier transform. Now let $\mathcal{G}(u,z)=\int_\mathbb{R}{\mathcal{F}(t,z)e^{uti}\,dt}$. Then \begin{align*} \frac{\partial\mathcal{G}(u,z)}{\partial z}&=\int_\mathbb{R}{\frac{\mathcal{F}(t,z)}{\partial z}e^{uti}\,dt} \\ &=\iint_{\mathbb{R}^2}{e^{upti}\mathcal{F}(pt,z)\cdot e^{u(1-p)ti}\mathcal{F}((1-p)t,z)\,\frac{d^2(p,t)}{t}} \\ &=\mathcal{G}(u,z)^2 \end{align*} by a change of variables. From the known value of $\mathcal{F}(t,0)$, $$\mathcal{G}(u,0)=\frac{1-e^{ui}(1-ui)}{u^2}$$ Despite the functional form, the singularity at $u=0$ is removable. Indeed, the value at $0$ has maximum modulus along the entire real line.
Solving the above ODE, $$\mathcal{G}(u,z)=\frac{1}{\frac{u^2}{1-e^{ui}(1-ui)}-z}=\sum_{n=0}^\infty{\left(\frac{1-e^{ui}(1-ui)}{u^2}\right)^{n+1}z^n}$$
Despite appearances, we're almost done. Let $$\alpha(u)=\int_1^\infty{\frac{e^{uti}}{t}\,dt}$$ You want \begin{align*} \mathbb{E}[X_n]&=\int_0^1{\mathbb{P}[X_n\leq t]\,dt}=\int_0^1{F_n\left(\frac{1}{t}\right)\,\frac{dt}{t}} \\ &=\int_1^\infty{F_n(u)\,\frac{du}{u}} \end{align*} which has generating function $$\mathcal{E}(z)=\int_1^\infty{\mathcal{F}(t,z)\,\frac{dt}{t}}=\frac{1}{2\pi}\int_\mathbb{R}{\overline{\alpha(u)}\mathcal{G}(u,z)\,du}$$ since the Fourier transform is an isometry.
Comparing coefficients of $z^n$, $$\mathbb{E}[X_n]=\frac{1}{2\pi}\int_\mathbb{R}{\int_1^\infty{\frac{e^{-uti}}{t}\left(\frac{1-e^{ui}(1-ui)}{u^2}\right)^{n+1}\,dt}\,du}$$ At large $n$, integral is dominated by the maximum near $0$; by Laplace's method \begin{align*} \mathbb{E}[X_n]&\approx\frac{1}{2\pi}\iint_{\mathbb{R}\times[1,\infty)}{\frac{e^{-uti}}{t}e^{(n+1)\left(-\ln(2)+4i\cdot\frac{u}{6}-\left(\frac{u}{6}\right)^2\right)}\,d^2(u,t)} \\ &=\frac{1}{2\pi}\int_1^\infty{\frac{3}{2^n}\sqrt{\frac{\pi}{1+n}}e^{-\frac{(3t-2(n+1))^2}{1+n}}\,\frac{dt}{t}} \\ &=\frac{3}{2^{n+1}\sqrt{\pi(n+1)}}\int_{1-2n}^\infty{\frac{e^{-\frac{u^2}{1+n}}}{u+2(n+1)}\,du} \end{align*}
Unfortunately, this decays exponentially in $n$, which other arguments have shown impossible.