$$I(X_{1:n};Y_{1:n})=\sum_{i=1}^n [H(Y_i|Y_{1:i-1})-H(Y_i,X_{i:n}|Y_{1:i-1},X_{1:i-1})+H(X_{i:n}|Y_{1:i-1},X_{1:i-1})]$$
Edited:
I know a little how to go further $$I(X_{1:n};Y_{1:n})=H(Y_{1:n})+H(Y_{1:n}|X_{1:n})$$ The first part of this, match the fist part of the result so we should prove this instead : $$H(Y_{1:n}|X_{1:n})=\sum_{i=1}^n [H(Y_i,X_{i:n}|Y_{1:i-1},X_{1:i-1})-H(X_{i:n}|Y_{1:i-1},X_{1:i-1})] $$
Starting with $$I(X_{1:n};Y_{1:n})=H(Y_{1:n})+H(Y_{1:n}\mid X_{1:n}) \tag{1}$$
For the first term we have (chain rule) :
$$H(Y_{1:n})=H(Y_n\mid Y_1 \cdots Y_{n-1})+H(Y_{n-1}\mid Y_1 \cdots Y_{n-2}) +\cdots =\sum_{i=1}^n H(Y_i\mid Y_{1,i-i}) \tag{2}$$
The second is equivalent, only that conditioned on $X_{1:n}$, then we can write $$H(Y_{1:n}\mid X_{1:n}) =\sum_{i=1}^n H(Y_i\mid Y_{1,i-i},X_{1:n}) \tag{3}$$
Now, because we want the get "up" the $X_{i:n}$ part, we split the multidimensional $X$ variable in two parts:
$$ H(Y_i\mid Y_{1,i-i},X_{1:n})=H(Y_i\mid Y_{1,i-i},X_{1:i-1},X_{i:n}) \tag{4}$$
Recalling $H(A \mid B) = H(A,B) - H(B)$, and $H(A \mid B,C) = H(A,B\mid C) - H(B\mid C)$ we get (taking $A=Y_i$, $B=X_{i:n}$, $C=(Y_{1,i-i},X_{1:i-1})$ ) :
$$ H(Y_i\mid Y_{1,i-i},X_{1:i-1},X_{i:n})= H(Y_i,X_{i:n}\mid Y_{1,i-i},X_{1:i-1})- H(X_{i:n}\mid Y_{1,i-i},X_{1:i-1}) \tag{5}$$
I leave to you the task of putting all together.