Explicit expression of an alternating polynomial in characteristic $2$? Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.
Consider variables $X=(X_1, \ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, \ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + \cdots + T_n$. 
A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, \ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$. 
It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial $V(X) = \prod_{i<j} (X_j-X_i)$, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$.
Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T_1 Y^{n-1} + \cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.    
That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even. 
I already figured out what is the "elementary" alternating polynomial $u/V$ and what is the Artin-Schreir equation it satisfies: 
$u(X) = \sum_{s \ {\rm is\ even}} X^{n-1}_{s(1)} \cdots X^0_{s(n)}$ and it satisfies the Artin-Schreier equation $X^2 + X = \frac{u(X) u(s_0 X)}{\Delta}$, where $s_0$ is any odd permutation (e.g., transposition), and $\Delta$ is again the discriminant. (Note that $u(X) + u(s_0 X) = V$.)
My question is: Does there exist a nice formula for $\frac{u(X) u(s_0 X)}{\Delta}$ in terms of $T$? 
 A: (Here is a more detailed version of Felipe's answer.)
In 1976 Elwyn Berlekamp defined characteristic 2 analogues of the discriminant and its square root, related to the expressions you wrote down.  Later these were related to what you get if you lift the original polynomial to characteristic 0, compute the usual discriminant, and reduce modulo 8; this should give you a formula, though maybe not a very nice one.  Here are the references:
Berlekamp, E. R. An analog to the discriminant over fields of characteristic two.  J. Algebra  38  (1976), no. 2, 315--317.  
Revoy, Philippe Discriminant d'une extension s�parable d'anneaux.  C. R. Acad. Sci. Paris S�r. I Math.  298  (1984),  no. 7, 123--126.
Berg�, A.-M.; Martinet, J. Formes quadratiques et extensions en caract�ristique $2$. (French) [Quadratic forms and extensions in characteristic $2$]  Ann. Inst. Fourier (Grenoble)  35  (1985),  no. 2, 57--77.
Wadsworth, Adrian R. Discriminants in characteristic two.  Linear and Multilinear Algebra  17  (1985),  no. 3-4, 235--263.
Waterhouse, William C. Discriminants of �tale algebras and related structures.  J. Reine Angew. Math.  379  (1987), 209--220.
A: Berlekamp, E. R.
An analog to the discriminant over fields of characteristic two.
J. Algebra 38 (1976), no. 2, 315--317. 
