I had a fairly useless comment posted earlier which I apologize for. It turns out much is known though. Below everything is restricted to zero applied field.
For a finite square grid of size $m$ by $n$ with periodic boundary conditions, the expression for the partition function $Z$ (from which the entropy per site can be worked out) was given already by Kaufmann (note that you can find freely accessible copies of this paper through google) very shortly after Onsager's solution:
$Z=\frac{1}{2}(2\sinh(2H))^{mn/2}\left\{\prod_{r=1}^n\left(2\cosh(\frac{m}{2}\gamma_{2r-1})\right)+\prod_{r=1}^n\left(2\sinh(\frac{m}{2}\gamma_{2r-1})\right)+\prod_{r=1}^n\left(2\cosh(\frac{m}{2}\gamma_{2r})\right)+\prod_{r=1}^n\left(2\sinh(\frac{m}{2}\gamma_{2r})\right)\right\}$
where, $H=J/k_BT$, $H'=J'/k_BT$, $\cosh\gamma_j=\cosh(2H^*)\cosh(2H')-\sinh(2H^*)\sinh(2H')\cos(\pi j/n)$ where $H^*$ is defined to be $tanh H^*:= \exp(-2H)$, with $J,J'$ being the coupling constants in the horizontal and vertical directions, respectively.$$ \begin{split} Z &=\frac{1}{2}\big(2\sinh(2H)\big)^{mn\over 2}\left\{\prod_{r=1}^n 2\cosh\Big(\frac{m}{2}\gamma_{2r-1}\Big)+\prod_{r=1}^n2\sinh\Big(\frac{m}{2}\gamma_{2r-1}\Big)\right. \\ &\quad\quad\qquad\qquad\qquad\qquad\left.+\prod_{r=1}^n2\cosh\Big(\frac{m}{2}\gamma_{2r}\Big)+\prod_{r=1}^n2\sinh\Big(\frac{m}{2}\gamma_{2r}\Big)\right\} \end{split} $$ where
- $H=J/k_BT$, $H'=J'/k_BT$, with $J,J'$ being the coupling constants in the horizontal and vertical directions, respectively,
- $\cosh\gamma_j=\cosh(2H^*)\cosh(2H')-\sinh(2H^*)\sinh(2H')\cos(\pi j/n)$ and
- $H^*$ is defined to as the solution to the equation $\tanh H^*:= \exp(-2H)$.
A fuller explanation is in chapter IV of McCoy and Wu's book "The Two-dimensional Ising model". Another way to derive the partition function which has proved useful for generalizations to antiperiodic boundary conditions was worked out in this paper by V.N. Plechko.
On an infinitely long cylinder with circumference $L$ I was able to find in a 2004 paper by Huang et al an expression for the free energy (with $J=J'=1$):
$f=-\frac{1}{2}\ln(4z)-\frac{1}{2L}\sum_p\int_0^{2\pi}\frac{d\phi}{2\pi}\ln\left[z+z^{-1}-\Phi_p(\phi)\right]$
where $$ f=-\frac{1}{2}\ln(4z)-\frac{1}{2L}\sum_p\int_0^{2\pi}\frac{d\phi}{2\pi}\ln\left[z+z^{-1}-\Phi_p(\phi)\right] $$ where now $z=\sinh(2\beta)$ and $\Phi_p(\phi)=\cos\phi+\cos\frac{2\pi p}{L}$.
- $z=\sinh(2\beta)$ and
- $\Phi_p(\phi)=\cos\phi+\cos\frac{2\pi p}{L}$.
