I do not know if the following is a "standard/correct/natural" method of defining the dual bimodule but here is what i have thought:  

Given an $A$-$B$-bimodule $M$ we can always view this as a right $B\otimes A^{op}$-module setting $m(b\otimes a^{op})=amb$. Conversely, if $M$ is a right  $B\otimes A^{op}$-module then we get an $A$-$B$-bimodule by setting $an=m(1\otimes a^{op})$ and $nb=n(b\otimes 1)$.  
 (Here $a^{op}$ stands for the element $a$ of the algebra $A$ viewed as an element of the opposite algebra/ring $A^{op}$). 

So, start with your $A$-$B$-bimodule, view it as a right $B\otimes A^{op}$-module and then get the [dual module][1] $Hom_{B\otimes A^{op}}(M,B\otimes A^{op})$. This will be a left $B\otimes A^{op}$-module. Following a strategy similar to the one mentioned in the preceding paragraph this can be viewed as an $A^{op}$-$B^{op}$-bimodule, which may be what you are looking for.  
(I will try to add more details asap). 


  [1]: https://en.wikipedia.org/wiki/Dual_module