Does a four-variable analog of the Hall-Witt identity exist?

$\renewcommand{\mod}[1]{~(\text{mod $#1$})} \newcommand{congr}{\equiv} $The last statement in your question (which is not a mere question) also suggests a solution of the more general problem, namely the $n$-variable analog of the Hall-Witt formula for any $n\geq 2$.

Let $x_1$, $\ldots$, $x_n$ be the variables. If $w=w(x_1,x_2,\ldots,x_n)$ is any word in $x_i$'s and their inverses, define the word $\gamma w$ by $(\gamma w)(x_1,x_2,\ldots,x_n):=w(x_2,\ldots,x_n,x_1)$, and set $W:=w(\gamma w)^{-1}$. Then $$W(\gamma W)\cdots(\gamma^{n-1}W)=1~.$$ The requirement that $W$ is nontrivial in each variable is easily satisfied. In this way you can produce $n$-variable analogs of the Hall-Witt formula by truckload.

To make things more interesting, you may require, say, that $W$ must be representable by an expression built from the variables and their inverses (the 'tokens') by making commutators $[u,v]$, where $u$ and $v$ are already built expressions, and by conjugations $u^y$, where $u$ is an already built expression, but not a token, and $y$ is a token; then the requirement is that $W=u$ for an expression built in this way that is not a single token. Now the things are no longer so simple. Am I far wrong in supposing that in fact you had some such additional conditions in mind when you formulated your question?

I have no idea how to go about finding an example for $n=4$ or proving that it does not exist. The only (rather weak) restriction on the word $w$ I have found so far is $l_1=l_2=\cdots=l_n$, where $l_i$ is the sum of the exponents of the appearances of $x_i^{\pm1}$ in $w$.

If we relax the condition imposed on $W$ and require only that $W\in[G,G]$, where $G$ is the free group generated by the variables $x_1$, $\ldots$, $x_n$, then we can give the full solution of the problem, since for every $w\in G$ we have $w(\gamma w)^{-1}\in[G,G]$ if and only if $l_1(w)=l_2(w)=\cdots=l_n(w)$. Here $\gamma$ is the automorphism of $G$ that sends $x_i$ to $x_{i+1}$ for $1\leq i<n$, and sends $x_n$ to $x_1$. We must also tell how the functions $l_i: G\to\mathbb{Z}$ are defined. Let $A$ be the free abelian (additive) group generated by $x_1$, $\ldots$, $x_n$, and let $h\colon G\to A$ be the homomorphism sending $x_i\in G$ to $x_i\in A$ for $1\leq i\leq n$; note that $\ker h=[G,G]$. For every $w\in G$ we have $h(w)=\sum_{i=1}^n l_i(w)x_i$: this defines the $l_i$'s.

For example, if $w=x_1^{-1}x_2^{-1}\cdots x_{n-1}^{-1}x_n^{-1}$, then $W=w(\gamma w)^{-1}=[x_1,x_nx_{n-1}\cdots x_2]$. When $n=3$ we obtain the identity $$ [x,zy][y,xz][z,yx]=1~, $$ which is a humble cousin of the Hall-Witt formula, and is probably quite useless (nice as it is).
If you wish you can rewrite $W=[x,zy]=x^{-1}y^{-1}z^{-1}xzy$ as a product of iterated commutators, $W=[x,y][x,z][[x,z],y]$, or perhaps in the well-known form $W=[x,y][x,z]^y$. (Mark that here $[x,y]=x^{-1}y^{-1}xy$; some authors define the commutator as $[x,y]:=xyx^{-1}y^{-1}$.) Note that the example $w=w_1(a,b,c,d)=a^{-1}bc^{-1}b^{-1}dad^{-1}$ provided by weux082690 has $l_a(w)=l_b(w)=l_d(w)=0$ but $l_c(w)=-1$, thus $w(\gamma w)^{-1}\notin[G,G]$.

By a sort of a retrograde progress, let us return to the beginning. We are going to prove the following:

Let $G$ be the free group with the free generators $x_1$, $\ldots$, $x_n$, $n\geq 2$, and let $\gamma$ be the automorphism of $G$ that rotates the generators one place to the right, that is, $\gamma x_i=x_{i+1}$ for $1\leq i<n$ and $\gamma x_n=x_1$. Then $g\in G$ has the property that $g(\gamma g)\cdots(\gamma^{n-1}g)=1$ iff there exists $h\in G$ such that $g=h(\gamma h)^{-1}$.

Proof. $~$The sufficiency is clear.

Necessity. Let $T$ be the set of 'tokens' $\{x_1,x_1^{-1},\ldots,x_n,x_n^{-1}\}$, and let $T^*$ denote the free monoid (of 'words') generated by $T$; the neutral element of $T^*$ is the empty word $\varepsilon$. Then $G=T^*/{\sim}$, where $\sim$ is the congruence on the monoid $T^*$ generated by $x_i^\alpha x_i^{-\alpha}\sim\varepsilon$, $1\leq i\leq n$, $\alpha=\pm1$. The equivalence class of the empty word is the multiplicative identity of the free group: $\varepsilon/{\sim}=1_G=1$. Every equivalence class $g\in G$ contains a unique reduced word $\varrho(g)$ which does not contain any pair of consecutive tokens that are inverses of each other. Given a word $w\in T^*$, we obtain the reduced word $\varrho(w/{\sim})$ by repeatedly applying the reductions $ux_i^{\alpha}x_i^{-\alpha}v\to uv$, $1\leq i\leq n$, $\alpha=\pm1$, $u,v\in T^*$; it is easy to verify that this system of reductions is locally confluent, thus it has the Church-Rosser property, and so there is in fact a unique reduced word in each equivalence class. The rotation $\gamma$ of generators induces the double rotation of tokens (the rotation of the tokens with the exponent $1$ and also the rotation of the tokens with the exponent $-1$); the automorphism of the free monoid $T^*$ determined by this double rotation we still denote by $\gamma$.
$\quad$Let $w=x_{i_1}^{\alpha_1}x_{i_2}^{\alpha_2}\cdots x_{i_m}^{\alpha_m}\in T^*$. We denote by $|w|$ the length $m$ of the word $w$. For each $i$, $1\leq i\leq n$, we denote by $l_i(w)$ the sum of the exponents of all tokens $x_i^{\pm1}$ appearing in $w$, and by $l(w)$ we denote the sum of all exponents $\alpha_1+\alpha_2+\cdots+\alpha_m=l_1(w)+\cdots+l_n(w)$. Always $|w|\congr l(w)\mod{2}$, because every term in $|w|-l(w)=(1-\alpha_1)+(1-\alpha_2)+\cdots+(1-\alpha_m)$ is ether $0$ or $2$. For each $i$, $1\leq i\leq n$, we have $l_i(uv)=l_i(u)+l_i(v)$ for all $u,v\in T^*$, and $l_i(w)$ is constant on every equivalence class $g\in G$.
$\quad$Now suppose that $g\in G$ has the property $g(\gamma g)\cdots(\gamma^{n-1} g)=1$, and let $w:=\varrho(g)$. Then $l_n(w)+l_n(\gamma w)+\cdots+l_n(\gamma^{n-1} w)=l_n(\varepsilon)=0$; since $l_n(\gamma w)=l_{n-1}(w)$, $\ldots$, $l_n(\gamma^{n-1} w)=l_1(w)$ we have $l(w)=l_1(w)+\cdots+l_n(w)=0$, therefore the word $w$ is of even length, $|w|=2k$. We split the word $w$ as $w=w_1w_2$, where $|w_1|=|w_2|=k$. We are assuming that $k>0$, since the case $k=0$ is trivial. The reduction process must reduce the word $$ W := w_1w_2(\gamma w_1)(\gamma w_2)(\gamma^2w_1)(\gamma^2w_2)\cdots (\gamma^{n-1}w_1)(\gamma^{n-1}w_2) $$ to the empty word. Since the words $w_1w_2$, $(\gamma w_1)(\gamma w_2)$, $\ldots$, $(\gamma^{n-1}w_1)(\gamma^{n-1}w_2)$ are reduced, the only places in the word $W$ where the reductions can be applied are at the points of contact between subwords $w_2$ and $\gamma w_1$, $\gamma w_2$ and $\gamma^2 w_1$, $\ldots\,$ Consider the effect of the reduction process on the subword $w_2(\gamma w_1)$ (mark that we can carry out the reductions in any order we choose, the result will be always the same). The first reduction eliminates some product $t\,t^{-1}$ from the center of $w_2(\gamma w_1)$, where $t$ is the last token in $w_2$ and $t^{-1}$ is the first token in $\gamma w_1$. We are left with $w_2'(\gamma w_1')$, where $|w_1'|=|w_2'|=k-1$. If $k>1$, there may be another reduction applicable at the center of $w_2'(\gamma w_1')$ (and nowhere else in this word), so we apply it, and so on. In fact the reductions must proceed to the bitter end, reducing the initial word $w_2(\gamma w_1)$ to the final empty word. For suppose that the reduction process stops after $r<k$ reductions; then the reduction process, applied to each of the subwords $(\gamma w_2)(\gamma^2 w_1)$, $\ldots$, $(\gamma^{n-2}w_2)(\gamma^{n-1}w_1)$, will likewise stop after $r$ reductions, and we will have a nonempty reduced word on our hands, which cannot be, because the word $W$ must reduce to the empty word. Let $h:=w_1/{\sim}$. Since $w_2(\gamma w_1)\sim\varepsilon$, it follows that $w_2/{\sim}=(\gamma h)^{-1}$, whence $g=(w_1/{\sim})(w_2/{\sim})=h(\gamma h)^{-1}$.$~$ Done.

I don't know how you would reformulate it in terms of commutators, but $W(a,b,c,d) = a^{-1}bc^{-1}b^{-1}da^{-1}d^{-1}ab^{-1}a^{-1}cdc^{-1}b$ works with $w_1(a,b,c,d) = a^{-1}bc^{-1}b^{-1}dad^{-1}$ and $w_2(a,b,c,d) = ab^{-1}a^{-1}cdc^{-1}b$. Therefore, $w_2(a,b,c,d)*w_1(b,c,d,a) = ab^{-1}a^{-1}cdc^{-1}b * b^{-1}cd^{-1}c^{-1}aba^{-1} = 1$

Update: by shuffling around the letters a bit, I got it in somewhat commutator terms: $w_1(a,b,c,d) = b^{-1}c^{-1}ba^{-1}dad^{-1} = (c^{-1})^b * [a, d^{-1}]$ and $w_2(a,b,c,d) = ab^{-1}a^{-1}bc^{-1}dc = [a^{-1}, b] * d^c$ so $W(a,b,c,d) = (c^{-1})^b * [a, d^{-1}] * [a^{-1}, b] * d^c$.