We need to revisit the definition of k-fold cross-validation: train set is divided into k equal parts. Each part is used for validation once. Cross-validation error is the average of k validation errors:
$CV(\lambda) = \frac{1}{k}\sum\limits_{i=1}^{k}E_i(\lambda)$
(For more details on cross-validation, read Hastie & Tibshirani, 2009)
As such, the standardization parameters (mean and standard deviation) for the test set should be the "average" of those used during cross-validation.
Let $\mu_i$'s $(i=1,2,...,k)$ be the means used during cross-validation. Then, the "average" mean used for standardizing the test set is:
$\mu_{avg} = \frac{1}{k}\sum\limits_{i=1}^k\mu_i
\\ = \frac{1}{k}\frac{(k-1)\sum\limits_{j=1}^{n}x_j}{\frac{n(k-1)}{k}} = \frac{1}{n}\sum\limits_{j=1}^{n}x_j$
which is mean of the train set.
It's tempting (and intuitive) to extrapolate this result and use the standard deviation of the whole train set to standardize the test set. However, the algebra for deriving $\sigma_{avg}$ is trickier so I haven't been able to prove it. Maybe someone can help me?