Let $F:\mathbb{S}^n\rightarrow \mathbb{S}^n$ be the antipodal map, then it's differential $D_pF:T_p\mathbb{S}^n\rightarrow T_{-p}\mathbb{S}^n$ is given by $X\mapsto -X$. Consider the volume form $dx^1\wedge \cdots \wedge dx^{n+1}$ on $\mathbb{R}^{n+1}$, and the vector field: $$X=x^i\frac{\partial}{\partial x^i}$$ on $\mathbb{R}^{n+1}$. Let $\iota:\mathbb{S}^n\rightarrow \mathbb{R}^{n+1}$ bet he inclusion, then: $$\omega=i^*(X\lrcorner\alpha)$$ is the standard volume form on $\mathbb{S}^n$. Let $v_1,\dots, v_n\in T_p\mathbb{S}^n$ be arbitrary, then: $$ \begin{align} (F^*\omega)_p(v_1,\dots, v_n)=&\omega_{-p}(-v_1,\dots, -v_n)\\ =&(-1)^n\alpha_{-p}(X_{-p},v_1,\dots, v_n) \end{align} $$ Now, I want to say that since the volume form is constant, we have $\alpha_p=\alpha_{-p}$, and that since $v_1,\dots, v_n$ originally lied $T_p\mathbb{S}^n$ they now lie in $T_p\mathbb{S}^n$ because I have factored out the negative signs. Thus we can write: $$ \begin{align} (F^*\omega)_p(v_1,\dots, v_n)=&(-1)^n\alpha_p(-X_p,v_1,\dots, v_n)\\ =&(-1)^{n+1}\omega_p(v_1,\dots,v_n) \end{align} $$ so $F^*\omega=(-1)^{n+1}\omega$. Does this make any sense? I always feel uneasy about pulling back my automorphisms...
Edit:
I agree it makes no sense to say that the volume form is constant. How about this:
We have that: $$X_{-p}=F_*(X_p)$$ where $F$ is regarded as the antipodal map on $\mathbb{R}^{n+1}$, hence: $$\begin{align} (F^*\omega)_p(v_1,\dots, v_n)=&\alpha_{-p}(F_*(X_p), -v_1,\dots, -v_n)\\ =&(F^*\alpha)_p(X_p,v_1,\dots, v_n) \end{align}$$ Now: $$(F^*\alpha)=d(x^1\circ F)\wedge \cdots d(x^{n+1}\circ F)=(-1)^{n+1}dx^1\wedge\cdots \wedge dx^{n+1}=(-1)^{n+1}\alpha$$ hence: $$(F^*\omega)_p(v_1,\dots, v_n) =(-1)^{n+1}\alpha_p(X_p,v_1,\dots, v_n)$$ implying that: $$(F^*\omega)=(-1)^{n+1}\omega$$ Does this work?