I wanted to comment on the question in order to ask more information, but apparently don't have enough points yet to comment, so will just "answer" your question.

A normal corridor variance swap spread would have payoff 

$$ 
\int_0^T (\sigma_t^X)^2 \theta(X_t - K_X) dt - \int_0^T (\sigma_t^Y)^2 \theta(Y_t - K_Y) dt 
$$

where $X_t$ is say the SX5E and $Y_t$ is the SPX, $K_X$ the lower corridor for SX5E and $K_Y$ the lower corridor for the SPX, and $\sigma_t^X$ resp. $\sigma_t^Y$ the vols of the two indices, $\theta$ is the Heaviside function. Note that I only have a lower corridor but could just have easily included an upper as well. Do you agree with the above payoff for a normal corr varswap spread?

The corr varswap spread you're trying to value on the other hand, has the following payoff:

$$ 
\int_0^T (\sigma_t^X)^2 \theta(X_t - K_X) dt - \int_0^T (\sigma_t^Y)^2 \theta(X_t - K_X) dt 
$$

Note that the term $\theta(X_t - K_X)$ occurs in both integrals now. Is my interpretation correct, before I continue trying to answer your question?

Following your comment below:

Well, I think this is a non-trivial problem in the sense that even though a normal corridor variance swap has an analytical replication expression, the 2nd integral above I'm not so sure of. In any case, I think you can see though why the price is cheaper than a normal corr varswap spread: If the SPX is highly negatively correlated to its volatility, it's reasonable to assume that the SX5E is less negatively correlated to the SPX vol. This means that the second integral above (the product of SPX vol with SX5E corridor) will have more value than a pure SPX corridor varswap. Hence the spread (both integrals above) will have less value.

I will continue thinking about an approximate analytical expression for the replication of the second integral above, but at this moment I don't see a straightforward answer.