I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't find any in complex dimension one, so let's look in dimension two. By classification of surfaces, the non algebraic surfaces would have Kodaira dimensions $\kappa=0$ or $1$. The $\kappa=0$ cases are, I believe, necessarily tori or K3's (there are some other examples, but these are either algebraic or non Kähler). As already noted by you and Francesco, the algebraic surfaces form a proper subset in moduli for these cases. The $\kappa=1$ surfaces are elliptic surfaces, and I expect that there should be plenty of non algebraic examples, although I don't have an example on hand.

Added later For my own reasons, I thought about this longer than I normally would. Here's an explicit example. Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma=\pi_1(C)$, and $\tilde C$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h:\Gamma\to E$. Define an action of $\Gamma$ on $\tilde C\times E$ by $g(x,y)= (gx, y+h(g))$, and let $S$ be the quotient.

$S$ is Kähler. If $h$ has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde C\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f:S\to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(g)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.

I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't find any in complex dimension one, so let's look in dimension two. By classification of surfaces, the non algebraic surfaces would have Kodaira dimensions $\kappa=0$ or $1$. The $\kappa=0$ cases are, I believe, necessarily tori or K3's (there are some other examples, but these are either algebraic or non Kähler). As already noted by you and Francesco, the algebraic surfaces form a proper subset in moduli for these cases. The $\kappa=1$ surfaces are elliptic surfaces, and I expect that there should be plenty of non algebraic examples, although I don't have an example on hand.

Added later For my own reasons, I thought about this longer than I normally would. Here's an explicit example. Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma=\pi_1(C)$, and $\tilde C$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h:\Gamma\to E$. Define an action of $\Gamma$ on $\tilde C\times E$ by $\gamma(x,y)= (\gamma x, y+h(\gamma))$, and let $S$ be the quotient.

$S$ is Kähler. If $h$ has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde C\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f:S\to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.

Additional remarks: This has $\kappa=1$ when $g>1$. When $g=1$, one can see, with a bit of thought, that $S$ is torus which contains an elliptic curve but it is not isogenous to a product (i.e. Poincar'e reducibility fails for tori).

I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't find any in complex dimension one, so let's look in dimension two. By classification of surfaces, the non algebraic surfaces would have Kodaira dimensions $\kappa=0$ or $1$. The $\kappa=0$ cases are, I believe, necessarily tori or K3's (there are some other examples, but these are either algebraic or non Kähler). As already noted by you and Francesco, the algebraic surfaces form a proper subset in moduli for these cases. The $\kappa=1$ surfaces are elliptic surfaces, and I expect that there should be plenty of non algebraic examples, although I don't have an example on hand.

I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't find any in complex dimension one, so let's look in dimension two. By classification of surfaces, the non algebraic surfaces would have Kodaira dimensions $\kappa=0$ or $1$. The $\kappa=0$ cases are, I believe, necessarily tori or K3's (there are some other examples, but these are either algebraic or non Kähler). As already noted by you and Francesco, the algebraic surfaces form a proper subset in moduli for these cases. The $\kappa=1$ surfaces are elliptic surfaces, and I expect that there should be plenty of non algebraic examples, although I don't have an example on hand.

Added later For my own reasons, I thought about this longer than I normally would. Here's an explicit example. Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma=\pi_1(C)$, and $\tilde C$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h:\Gamma\to E$. Define an action of $\Gamma$ on $\tilde C\times E$ by $g(x,y)= (gx, y+h(g))$, and let $S$ be the quotient.

$S$ is Kähler. If $h$ has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde C\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f:S\to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(g)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.