The effect of $\lambda$ in the ridge regression estimator is that it "inflates" singular values $s_i$ of $X$ via terms like  $(s^2_i+\lambda)/s_i$. Specifically, if SVD of the design matrix is $X=USV^\top$, then $$\hat\beta_\mathrm{ridge} = V^\top \frac{S}{S^2+\lambda I} U y.$$
This is explained multiple times on our website, see e.g. @whuber's detailed exposition here: https://stats.stackexchange.com/questions/220243.


This suggests that selecting $\lambda$ much larger than $s_\mathrm{max}^2$ will shrink everything very strongly. I suspect that $$\lambda=\|X\|_2^2=\sum s_i^2$$ will be too big for all practical purposes.

I usually normalize my lambdas by the squared Frobenius norm of $X$ and have a cross-validation grid that goes from $0$ to $1$ (on a log scale).

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Having said that, no value of lambda can be seen as truly "maximum", in contrast to the lasso case. Imagine that predictors are *exactly* orthogonal to the response, i.e. that the true $\beta=0$. Any finite value of $\lambda<\infty $ for any finite value of sample size $n$ will yield $\hat \beta \ne 0$ and hence could benefit from stronger shrinkage.