By the corollary to theorem 4 in Kaimanovich's two page paper "Brownian motion on foliations: Entropy, Invariant measures, mixing" any foliation by leaves of sub-exponential volume growth must have a transverse invariant measure. An example of this is the foliation of the unit tangent bundle of $d$-dimensional hyperbolic space by the normal vectors to each horocycle (i.e. the strong stable foliation of the geodesic flow).

There is a $3$ dimensional manifold foliated by $2$-dimensional leaves without any transverse invariant measures which I think is due to Hirsch (I like it because because one can visualize it in Euclidean $3$ space easily).

To construct it take a solid torus $T = D \times S$ ($D$ a disk and $S$ the unit circle in the complex plane) and a solenoid map $f: T \to T$ which we take to be $s \mapsto s^2$ in the $S$ coordinate. Consider the foliation of $T \setminus f(T)$ by sets of the form $D \times \{s\}$ (each of these is a disk minus two smaller disks). Then identify the outer and inner boundaries of $T \setminus f(T)$ using $f$.

You get a compact foliated manifold (without boundary). Depending on whether the the $S$ coordinate of a point is eventually periodic under $s \mapsto s^2$ or not, the leaf the point belongs to can look like a $2$-sphere minus a Cantor set or a Torus minus a Cantor set. All leaves are dense.

The pasting you did makes is so that each time you want to go through a boundary torus you either square or take a square root of the $S$ coordinate.

In particular the holonomy along any curve which crosses the outer torus exactly once from the inside towards the outside is $s \mapsto s^2$ in the $S$ coordinate. Using this fact you can show there is no transverse invariant measure.