Yes, but only if you know the mean and variance (or standard deviation) of the signal. The maximum correlation will be the (variance of the signal + the mean-squared) times the total number of samples. If you do not already know the variance, calculating it would be similar to calculating the auto-correlation in terms of processing so nothing is really saved in that case. But in some cases you may have prior knowledge or ability to more easily estimate the variance (from a smaller sample set for example if the process is white and stationary).
To check my logic, consider what is involved to calculate correlation for a sampled signal when it is aligned with itself (for the maximum autocorrelation) compared to calculating the variance for the signal:
To correlate two signals: You multiply the two signals sample by sample, and then sum the result.
The equation to calculate the variance is similar, but you must subtract the mean from each sample before doing the product, and then you must divide by the total number of samples. (Important to note, this is not the unbiased estimate of the variance which scales by N-1, but the calculation for the 2nd Moment about the mean, which scales by N). The variance therefore is just a normalized autocorrelation at lag=0 with the mean removed.
To put very simply, CorrMax is just 1/N * the second moment. (The second moment is the average of the squares or mean-squared value). Variance is the 2nd moment with the mean removed (about the mean), so we have to add it back in as we've done above if we were starting with that particular factor known. However consider if we already know the mean-squared value, (or the square of the rms value since rms is commonly used), then the relationship is the mean-squared value times the number of samples.
Therefore in summary:
To determine the maximum correlation for a sequence X that is N samples long, assuming you already have either the mean and variance (2nd moment about the mean), use:
$$ CorrMax(X) = N(\sigma_X+\mu_X^2)$$
Where:
X is a vector of length N
$ \sigma_X $ is the second moment of X about the mean
$ \mu_X $ is the mean of X
Or alternatively calculated from the mean-squared value for the sequence,
$$ CorrMax(X) = N(X_{rms}^2)$$
Where:
X is a vector of length N
$ X_{rms}^2 $ is the second moment of X (mean-square of X)