Short answer:
On axis or central lobe field increases with the area of the slit (and thus the intensity $I\propto area^2$).
Long answer:
If you have a monochromatic plane wave traveling along the $z$ axis incident on
a diffraction slit of width $w$ (as shown in the image below), then the initial
field is $E(r,t) = E_0 e^{ik_0 z- \omega t}$. Now immediately after the slit
(assumed to be at $z=0$) the field is (ignoring the temporal component since
this is a monochromatic wave)
$$E(r) = E_0\operatorname{rect}(x/w),$$
where $\operatorname{rect}$ is the rectangle function (equal to one if
$|x/w|<1/2$ and zero outside this range).
Now we can still write the field as a sum of plane waves in what is known as
the angular spectrum representation, i.e.
$$E(r) = \int \tilde E(k_x) e^{ik_x x} dk_x/2\pi,$$
which we can Fourier transform to find the amplitude in the plane wave $\tilde
E(k_x)$, i.e.
$$\tilde E(k_x) = \int E(x) e^{ik_x x} dx.$$
The image above shows a sample of three of these plane wave components (at
$\theta = 0$ and at $\pm 45$ degrees).
Now if $E(x)$ has a finite width (e.g. by a slit of width $w$), then we can
change variables to $dx = wdx'$ and we see that this equation is proportional
to the width $w$. In particular, if the slit is a hard slit given by the
rectangle function, then we know that the Fourier transform is simply given by
$$\tilde E(k_x) = wE_0\operatorname{sinc}(wk_x/2\pi),$$
where $\operatorname{sinc}$ is the sinc function For the on axis (central
lobe) intensity, this gives
$$|\tilde E(k_x=0)|^2 = w^2|E_0|^2.$$
Note for a 2D "slit" or hole, the field for $k_x = k_y = 0$ would be
proportional to $\int dxdy \to area$, and thus the intensity would be
$|E_0|^2\times area^2$.
