Yes, here is the proof.
$\alpha\wedge{*}\beta=<\alpha,\beta>\Omega_{g}$, this is the definition of hodge star.
Pullback by $\varphi$, we get:
$LHS=\varphi^{*}(\alpha\wedge{*_{g}}\beta)=\varphi^{*}\alpha\wedge\varphi^{*}{*_{g}}\beta$
$RHS=\varphi^{*}(<\alpha,\beta>\Omega_{g})=\varphi^{*}<\alpha,\beta>\Omega_{\varphi^{*}g})=<\varphi^{*}\alpha,\varphi^{*}\beta>\Omega_{\varphi^{*}g}=\varphi^{*}\alpha\wedge*_{\varphi^{*}g}\varphi^{*}\beta$.
Since $\alpha,\beta$ are arbitrarily given, compare both sides we get the conclusion.
The answer is no. To see why, note that every self-dual $n$-form actually satisfies a stronger pointwise condition: If $\omega$ is self-dual with respect to some Riemannian metric $g$ and choice of orientation, then $\omega \wedge \omega = \omega \wedge *\omega = \langle \omega,\omega\rangle_g dV_g$ everywhere on $X$, where $dV_g$ is the Riemannian volume form and $\langle \cdot,\cdot\rangle_g$ is the pointwise inner product on $n$-forms induced by the metric. Thus a necessary condition for $\omega$ to be self-dual with respect to some metric is that $\omega\wedge\omega$ must be a nonnegative function times a positively oriented volume form. (An analogous statement holds for anti-self-dual forms.)
To see that your original condition is not sufficient, suppose $X$ is a $2n$-dimensional oriented Riemannian manifold with $n$ even, let $\alpha$ be a nontrivial $n$-form such that $\alpha\wedge\alpha \equiv 0$ -- for example, you could choose $\alpha$ to be a wedge of basis $1$-forms in local coordinates, extended to a global form by multiplying by a bump function. It follows also that
$$
*\alpha\wedge *\alpha = \langle *\alpha,\alpha\rangle_g\,dV_g = \langle \alpha,*\alpha\rangle_g\,dV_g = \alpha\wedge\alpha\equiv 0.
$$
Let $f$ be an arbitrary real-valued $C^\infty$ function on $X$, and let $\omega = \alpha + f{*\alpha}$. Then
$$\int_X\omega\wedge\omega = \int_X 2f\alpha\wedge *\alpha = \int_X 2f \langle\alpha,\alpha\rangle_g dV_g. $$
It is an easy matter to choose $f$ in such a way that it attains both positive and negative values, but such that the integral above is positive. The resulting $\omega$ satisfies your condition but is not self-dual with respect to any metric.
(The requirement that $n$ be even is so that $\alpha \wedge *\alpha = *\alpha\wedge\alpha$. If $n$ is odd, then $\omega\wedge\omega=0$ for every $n$-form $\omega$, and the entire question is moot.)
This suggests a more interesting question: On an oriented $2n$-manifold $X$, if $\omega$ is an $n$-form such that $\omega\wedge\omega$ is a nonnegative function times a positively oriented volume form, is there a Riemannian metric with respect to which $\omega$ is self-dual? I don't know the answer.
P.S. It seems to me that "norm" is an awfully inappropriate term for the quantity $\int_X \xi\wedge\xi$, because it's not positive or even nondegenerate, and there are many nontrivial $n$-forms for which it's zero. (Every decomposable form, for example.)
Best Answer
Quickly:
Assume you are dealing with closed oriented (simply-connected?) 4-manifold $M$.
There is an intersection form $Q_M$ on $H^2(M;\mathbb{Z})=H^2_{sing}(M;\mathbb{Z})$ given by $Q_M(\alpha,\beta)=(\alpha\smile\beta)([M])\in\mathbb{Z}$ which is symmetric for dimension reasons. Under de Rham isomorphism, $Q_M(\alpha,\beta)=\int_M\alpha\wedge\beta$. Since $\mathbb{Z}$ is torsion-free, $Q_M$ kills torsion elements so we can use $Q_M$ on $H^2_{sing}(M;\mathbb{R})\cong H^2_{dR}(M;\mathbb{R})$. Under Hodge you get harmonic 2-forms $\mathcal{H}^2(M)$.
Diagonalizing the form $Q_M$ over $\mathbb{R}$ gives a set of positive and negative eigenvalues (counted with multiplicities, by the corresponding set of eigenvectors in $H^2_{dR}(M;\mathbb{R})$). We define $b_2^+$ (resp. $b_2^-$) to be the number of positive (resp. negative) eigenvalues. We can follow this construction onto $\mathcal{H}^2(M)$ and get $b^\pm$. But since $Q_M$ is the same quadratic form on $H^2(M;\mathbb{R})$ and $\mathcal{H}^2(M)$, we have $b_i^\pm=b^\pm$, and so the signature $\sigma(M)=b^+-b^-$ is purely topological.