Sum of identically distributed but not independent Bernoulli's is non-uniform

First, some notations: for every $n$, let $X_{1:n}=(X_1,X_2,\ldots,X_n)$, and, for every word $w=(w_1,w_2,\ldots,w_n)$ in $\{0,1\}^n$, let $|w|=w_1+w_2+\cdots+w_n$. Now, to the proof, which uses crucially the exchangeability hypothesis.

Assume that, for some $n\geqslant1$, $|X_{1:n+1}|$ is uniformly distributed on $\{0,1,\ldots,n+1\}$.

Then, by exchangeability, for every word $w$ in $\{0,1\}^{n+1}$, $P(X_{1:n+1}=w)$ depends only on $|w|$. For each $k$ in $\{0,1,\ldots,n+1\}$, there are ${n+1\choose k}$ words $w$ in $\{0,1\}^{n+1}$ such that $|w|=k$, hence, for every word $w$ in $\{0,1\}^{n+1}$ such that $|w|=k$, $$P(X_{1:n+1}=w)={n+1\choose k}^{-1}P(|X_{1:n+1}|=k)=(n+2)^{-1}{n+1\choose k}^{-1}$$ For every word $v$ in $\{0,1\}^n$, the event $[X_{1:n}=v]$ is the disjoint union of the events $[X_{1:n}=v,X_{n+1}=0]$ and $[X_{1:n}=v,X_{n+1}=1]$. If $|v|=k$ for some $k$ in $\{0,1,\ldots,n\}$, then $|v0|=k$ and $|v1|=k+1$, hence $$P(X_{1:n}=v)=(n+2)^{-1}{n+1\choose k}^{-1}+(n+2)^{-1}{n+1\choose k+1}^{-1}$$ Now, it happens that $$(n+2)^{-1}{n+1\choose k}^{-1}+(n+2)^{-1}{n+1\choose k+1}^{-1}=(n+1)^{-1}{n\choose k}^{-1}\tag{$\ast$}$$ hence $$P(X_{1:n}=v)=(n+1)^{-1}{n\choose k}^{-1}$$ Summing these over the ${n\choose k}$ words $v$ in $\{0,1\}^n$ such that $|v|=k$, one gets, for every $k$ in $\{0,1,\ldots,n\}$, $$P(|X_{1:n}|=k)=(n+1)^{-1}$$ Thus, if $|X_{1:n+1}|=X_1+X_2+\cdots+X_{n+1}$ is uniformly distributed on $\{0,1,\ldots,n+1\}$, then $|X_{1:n}|=X_1+X_2+\cdots+X_n$ is uniformly distributed on $\{0,1,\ldots,n\}$.

By contraposition, this proves the desired statement -- and also that, as soon as $|X_{1:n}|=X_1+X_2+\cdots+X_{n}$ is uniformly distributed on $\{0,1,\ldots,n\}$ for some $n\geqslant1$, then $|X_{1:1}|=X_1$ is uniformly distributed on $\{0,1\}$, and, again by exchangeability, every $X_n$ is uniformly distributed on $\{0,1\}$, that is, necessarily, $p=\frac12$.

Exercise: Prove $(\ast)$.