Since this question doesn't have an answer yet, let me try to say something.
I'm not sure how helpful it is to talk about the "correct" definition of Calabi–Yau: different people use different definitions depending on the context, and the important thing is just to be clear about which definition you're using. For example, in birational geometry and minimal model theory, often the main thing one cares about is the numerical properties of the canonical bundle, so it makes sense to allow Calabi–Yau to mean any variety such that $K_X \equiv 0$. By contrast, from the point of view of differential geometry (about which I don't really know anything), one wants to understand the holonomy of a manifold, and then the condition that $h^i(\mathcal{O}_X)=0$ (which for varieties with trivial canonical bundle is, as I understand it, equivalent to the holonomy being exactly $SU(n)$ rather than a proper subgroup) is natural.
Something else that seems relevant to your question is the Beauville–Bogomolov decomposition theorem: this says if $X$ is a compact Kähler manifold with $K_X \equiv 0$, then there is a finite étale cover $\tilde{X} \rightarrow X$ such that $\tilde{X}$ is a product of Calabi–Yau manifolds (in the stricter sense), complex tori, and so-called hyperkähler manifolds. The point is that this shows that strict Calabi–Yaus are still something natural — they are one of the basic building blocks of $K$-trivial Kähler manifolds.
Examples: Of course what counts as an example depends on your definition. If we're allowing the most general definition, then as well as strict Calabi–Yaus and abelian varieties as you mentioned, we also have hyperkähler varieties as mentioned above. The Beauville–Bogomolov theorem then shows that these are (up to finite étale covers and products) all examples.
Examples of strict Calabi–Yaus:
Smooth hypersurfaces of degree $n+1$ in $\mathbf{P}^n$, for $n \geq 3$, including quintic threefolds. Adjunction shows that these have trivial canonical bundle, and the Lefschetz hyperplane theorem shows that the appropriate $h^i(\mathcal{O}_X)$ vanish. More generally, complete intersections of type $(n_1,\ldots,n_k)$ and dimension at least 2 in $\mathbf{P}^n$, where $n_1+\cdots+n_k=n+1$.
More generally, appropriate complete intersections in weighted projective spaces, or in products of projective spaces, give more examples.
I don't know many more. One interesting class of examples I can think of is elliptic Calabi–Yau threefolds, which means those possessing a map to a surface whose general fibre is an elliptic curve. Grassi–Morrison's fibre products of rational elliptic surfaces give some interesting examples here. Similarly, there are Calabi–Yaus threefolds with abelian surface fibrations: here an example is Horrocks–Mumford quintics.
Finally, let me say a little about
Examples of hyperkähler varieties:
- Very few are known. See this MO question.
For ease of notation, I'll explain how to construct the nowhere-vanishing holomorphic volume form in the case $n = 2$, where the Calabi-Yau hypersurface is a quartic K3 surface. This method generalises for all $n$.
Let's first focus on the affine patch:
$$ U = \{ [1 : z_1 : z_2 : z_3] \ \vert \ (z_1, z_2, z_3) \in \mathbb C^3 \} \subset \mathbb{CP}^3.$$
On this affine patch $U$, the hypersurface is defined by a polynomial equation:
$$ f_{\rm quartic}(z_1, z_2, z_3) = 0.$$
For each $i \in \{1,2,3 \}$, we define the open subset
$$ V_i = \left\{ (z_1, z_2, z_3) \in \mathbb C^3 \ \big\vert \ \frac{\partial f_{\rm quartic}}{\partial z_i}(z_1, z_2, z_3) \neq 0\right\} \subset U.$$
Since $f_{\rm quartic}(z_1, z_2, z_3)$ is non-singular, we have
$$ U = V_1 \cup V_2, \cup V_3.$$
Now, observe that $z_2$ and $z_3$ form a set of holomorphic coordinates on $V_1$, by the holomorphic implicit function theorem (with $z_1$ determined holomorphically in terms of $z_2$ and $z_3$). Similar statements hold for $V_2$ and $V_3$.
We can therefore define non-vanishing holomorphic two-forms $\Omega_{V_1}$, $\Omega_{V_2}$ and $\Omega_{V_3}$ on $V_1, V_2, V_3$ respectively:
$$ \Omega_{V_1} = \frac{dz_2 \wedge dz_3}{\partial f_{\rm quartic}/\partial z_1}, \ \ \ \ \
\Omega_{V_2} = \frac{dz_3 \wedge dz_1}{\partial f_{\rm quartic}/\partial z_2}, \ \ \ \ \
\Omega_{V_3} = \frac{dz_1 \wedge dz_2}{\partial f_{\rm quartic}/\partial z_3},$$
It is easy to see that $$\Omega_{V_1} = \Omega_{V_2}$$ on $V_1 \cap V_2$. [One can verify this by multiplying the equation
$$ \frac{\partial f_{\rm quartic}}{\partial z_1} dz_1 + \frac{\partial f_{\rm quartic}}{\partial z_2} dz_2 + \frac{\partial f_{\rm quartic}}{\partial z_3} dz_3 = 0$$
by $\wedge dz_3$.]
Similarly, $\Omega_{V_2} = \Omega_{V_3}$ on $V_2 \cap V_3$, and $\Omega_{V_3} = \Omega_{V_1}$ on $V_3 \cap V_1$.
So $\Omega_{V_1}$, $\Omega_{V_2}$ and $\Omega_{V_3}$ glue together to define a non-vanishing holomorphic two-form $\Omega$ on the whole of the affine patch $U$.
There remains a possibility that $\Omega$ may vanish, or diverge, on the hyperplane
$$ H = \{ [0 : x_1 : x_2 : x_3] \ \vert \ [x_1 : x_2 : x_3] \in \mathbb{CP}^2 \}$$
which is not covered by the affine patch $U$. We must show that this does not happen.
Expressing this in the language of divisors, we know that ${\rm div}(\Omega) = nH$ for some $n \in \mathbb Z$, and our task is to show that $n = 0$. But this is obvious: since $\Omega$ is a meromorphic section of the canonical bundle, which is trivial for the quartic by adjunction, the divisor class ${\rm div}(\Omega)$ is the trivial divisor, hence $n = 0$. [Or if you want to avoid divisors, then just compute what $\Omega$ is after the change of coordinates and you'll see...]
As Gunnar pointed out, finding a Ricci-flat metric on the quartic K3 is a much more difficult problem. As far as I'm aware, my colleagues in string theory only know how to approximate this numerically.
Best Answer
As $X$ is a complete intersection, it follows from the Lefschetz Hyperplane Theorem that $\pi_1(X) \cong \pi_1(\operatorname{Gr}(2, 7)) = 0$, so $H^1(X; \mathbb{C}) = 0$. Since $X$ is Kähler, we have $H^1(X; \mathbb{C}) \cong H^{1,0}_{\bar{\partial}}(X)\oplus H^{0,1}_{\bar{\partial}}(X)$ and hence $H^{0,1}_{\bar{\partial}}(X) = 0$. Finally, by Dolbeault's theorem, we see that $H^1(X, \mathcal{O}_X) \cong H^{0,1}_{\bar{\partial}}(X) = 0$.