Sorry, I got confused there for a minute. The answer is in fact just "yes". I'm not sure why this was voted down.
The reason is that the border rank of 2x2 matrix multiplication is seven. This was proved 5 years ago by Landsberg, though it took quite some effort for me to track this down. Sometimes "yes" is the best one can do at short notice. Perhaps individuals who enjoy taking points away from people might consider this!
However, Winograd's algorithm can do nxn matrix multiplication with n^3/2 + 2n^2 multiplications (and many more additions than usual). This is only useful if the coefficients are very large because of the cost of the extra additions. Moreover, this is not a block algorithm, i.e. it cannot be applied recursively. Anyhow, for 2x2 matrices, n = 2 and so the total number of multiplications is 2^3/2 + 2n^2 = 12 which is more than the usual 8 multiplications (or 7 if you use Strassen). Similarly for n = 3 it is less efficient than the naive 27 multiplications (and even less efficient than Lederman's algorithm: 23 multiplications and Bard's algorithm: 22 multiplications), but for n = 4 we get 4^3/2 + 2x4^2 = 64 which is the same as the naive algorithm and for n = 6 we get 6^3/2 + 2x6^2 = 180 which is well under the usual 216 multiplications required by the naive algorithm and even less than the number of multiplications required if you use one level of Strassen. Already by n = 8 Strassen is better if you go by the number of multiplications alone. Of course in practice the coefficients need to be really large (perhaps tens of machine words) before either Winograd or Strassen are better than the naive algorithm.
Note that if you break a larger matrix up into 2x2, i.e. 4 blocks then the smaller submatrices no longer commute (matrix multiplication is not commutative). This is why Winograd's algorithm is not a block algorithm like the Strassen method.
The reason is that the border rank of 2x2 matrix multiplication is seven. This was proved 5 years ago by Landsberg, though it took quite some effort for me to track this down. Sometimes "yes" is the best one can do at short notice. Perhaps individuals who enjoy taking points away from people might consider this!
However, Winograd's algorithm can do nxn matrix multiplication with n^3/2 + 2n^2 multiplications (and many more additions than usual). This is only useful if the coefficients are very large because of the cost of the extra additions. Moreover, this is not a block algorithm, i.e. it cannot be applied recursively. Anyhow, for 2x2 matrices, n = 2 and so the total number of multiplications is 2^3/2 + 2n^2 = 12 which is more than the usual 8 multiplications (or 7 if you use Strassen). Similarly for n = 3 it is less efficient than the naive 27 multiplications (and even less efficient than Lederman's algorithm: 23 multiplications and Bard's algorithm: 22 multiplications), but for n = 4 we get 4^3/2 + 2x4^2 = 64 which is the same as the naive algorithm and for n = 6 we get 6^3/2 + 2x6^2 = 180 which is well under the usual 216 multiplications required by the naive algorithm and even less than the number of multiplications required if you use one level of Strassen. Already by n = 8 Strassen is better if you go by the number of multiplications alone. Of course in practice the coefficients need to be really large (perhaps tens of machine words) before either Winograd or Strassen are better than the naive algorithm.
Note that if you break a larger matrix up into 2x2, i.e. 4 blocks then the smaller submatrices no longer commute (matrix multiplication is not commutative). This is why Winograd's algorithm is not a block algorithm like the Strassen method.