As this is a representation theory question, the connection to affine Hecke algebras deserves a few more words. The (degenerate) affine Hecke algebra, $H_d$ is isomorphic as a vector space to $\mathbb{C}[x_1,\ldots,x_d]\otimes\mathbb{C}S_d$, so that $\mathbb{C}[x_1,\ldots,x_d]\otimes 1$ and $1\otimes\mathbb{C}S_d$ are subalgebras (which we identify with $\mathbb{C}[x_1,\ldots,x_d]$ and $\mathbb{C}S_d$ respectively) and satisfy the mixed relations are $x_js_i=s_ix_j$ if $j\neq i,i+1$, and $x_{i+1}s_i=s_ix_i+1$. There is an equivalent story in the nondegenerate case as well.

Now, the subalgebra $\mathbb{C}[x_1,\ldots,x_d]$ is maximal commutative, and given a finite dimensional $H_d$-module $M$, the Chinese remainder theorem yields a decomposition $$M=\bigoplus_{\underline{a}\in\mathbb{C}^n}M_{\underline{a}}^{\mathrm{gen}},$$ where $M_{\underline{a}}^{\mathrm{gen}}$ is the generalized $\underline{a}$-eigenspace for the action of $x_1,\ldots,x_d$ on $M$. That is, for $\underline{a}=(a_1,\ldots,a_d)$, there exists sufficiently large $N$ such that $(x_i-a_i)^N$ acts as 0 on $M_{\underline{a}}^{\mathrm{gen}}$ for all $i$. Let $M_{\underline{a}}\subset M_{\underline{a}}^{\mathrm{gen}}$ be the *honest* simultaneous eigenspace for the action of $x_1,\ldots,x_d$ (i.e. $x_i-a_i$ acts as 0).

First of all, it is straightforward to show that one can reduce the study of finite dimensional $H_d$-modules to so-called "integral" finite dimensional $H_d$-modules. These are the modules for which $M_{\underline{a}}^{\mathrm{gen}}\neq0$ implies $\underline{a}\in\mathbb{Z}^d$ and it is best to think of them as being "dense" among all the finite dimensional representations. Now, among the integral finite dimensional $H_d$-modules are the "calibrated" representations which satisfy $M_{\underline{i}}^{\mathrm{gen}}=M_{\underline{i}}$ for all $\underline{i}\in\mathbb{Z}^d$. These are parameterized *precisely* by skew tableaux and, on restriction to $\mathbb{C}S_d$, their characters are the skew Schur functions.

It is relatively easy to give a construction of these representations. Indeed, given a skew shape $\lambda/\mu$ consisting of $d$ boxes (and a "charge" $c$ which I will ignore for brevity), let $L(\lambda/\mu)$ be the $\mathbb{C}$ vector space with basis $$\lbrace v_T|T\mbox{ a standard filling of }\lambda/\mu\rbrace.$$ Define an action of $\mathbb{C}[x_1,\ldots,x_d]$ on $L(\lambda/\mu)$ by $x_i v_T=c(T_i)v_T$, where $c(T_i)$ denotes the content of the box in $T$ labeled $i$. In other words, $\mathbb{C}v_T=L(\lambda/\mu)_{(c(T_1),\ldots,c(T_d))}$.

The action of the symmetric group takes a bit more work. There is a natural action of $S_d$ on $\mathbb{Z}^d$ by place permutation, and similarly on the set of standard $\lambda/\mu$ tableaux. On the other hand, the mixed relation from the first paragraph implies that $s_jM_{\underline{i}}\subset M_{s_j\underline{i}}+M_{\underline{i}}$ for any integral finite dimensional $H_d$-module $M$, so $S_d$ cannot act on $L(\lambda/\mu)$ by permuting the entries in the tableaux $T$. To find the real action, we consider a system of intertwining operators $\phi_i=x_{i+1}s_i-s_ix_{i+1}=s_i(x_i-x_{i+1})+1$, $i=1,\ldots,d-1$. One can readily calculate that these operators satisfy the type $A$ braid relations and that $\phi_ix_j=x_{s_i(j)}\phi_i$. In particular, $\phi_jM_{\underline{i}}\subseteq M_{s_j\underline{i}}$. Note, however that $\phi_i^2=(x_i-x_{i+1})^2-1$ which may act as 0.

Since $L(\lambda/\mu)$ is calibrated, we must have $\phi_iv_T=Y_{i,T}v_{s_iT}$ for some scalars $Y_{i,T}$.
It is clear that we must have $Y_{i,T}=0$ if $s_iT$ is not standard. Using the formula for $\phi_i$ gives that $$(c(T_{i+1})-c(T_{i})s_iv_T=v_T-Y_{i,T}v_{s_iT}.$$ If the left-hand-side is 0, then $T_i$ and $T_{i+1}$ lie on the same diagonals, which cannot happen in a standard tableau. Therefore, in all cases we can solve for $s_iv_T$. Comparing the calculation $$s_iv_{s_iT}=(c(T_i)-c(T_{i+1})^{-1}(v_{s_iT}-Y_{i,s_iT}v_T)$$ to that for $s_iv_T$ suggests that $Y_{i,T}$ and $Y_{i,s_iT}$ should be related. In fact, the right guess is that they are the same. A quick calculation using this guess and the quadratic relation $s_i^2=1$ gives that $$Y_{i,T}=\pm\sqrt{(c(i+1)-c(i))^{-2}-1}.$$ It remains to check the braid relations, which is a bit tedious. This will settle the sign issue ($\pm=+$). There is a nice paper of Arun Ram, "Skew shape representations are irreducible" (arxiv:math.RT/0401326) which does this.

In a recent work of Kleshchev and Ram (arxiv:0809.0557), the authors describe a new model for the combinatorics above in the wider context of Quiver Hecke Algebras (aka Khovanov-Lauda-Rouquier algebras) which are connected to categorified quantum groups. Toward the end of the article is a nice (albeit brief) discussion of the connection to skew Schur functions.

with no common side of a box, then you just get the product of the Schur functions for the two pieces. So $s_{(3,1,1)/(2,1)}$ would be $e_1^2$. – Ben Webster♦ Apr 25 '10 at 16:24