Shannon information entropy value of 64 bit data?
I guess that the question's data processing

Uses &quot;64 bit binary value&quot; as symbol of the plaintext.
Enciphers these  with a 64-bit block cipher in ECB mode, and considers the symbols in the encrypted data flow to be the individual outputs of that block cipher.
Assimilates the observed/actual frequency of a symbol $i$ in an experimental data set to the probability $p_i$ it has for the source of that data set. Which of course is an approximation (argument: if we repeat the experiment we'll likely get different frequencies, even if the source is unchanged, thus the probabilities unchanged).
Applies the formula $\displaystyle H\ =\ \sum_{i\text{ with }p_i\ne0}p_i\,\log_2\left(\frac1p_i\right)$ for the per-symbol Shannon entropy of a source of independent symbols of known probabilities $p_i$, even though nothing in the question supports that hypothesis of independence of the symbols produced by the source.

Assuming the above, reversibility of the transformation made on the symbols implies that encryption leaves $H$ exactly unchanged. The $p_i$ summed are the same, only for different $i$. The discrepancy between the reported 15.497477 before encryption and 15.0525 after could come from basing the computation of the second value on output that is not the encryption of the input used for the computation of the first value.
And independently, what's computed is something about the data (I don't know a proper name) produced by a source, but we have no assurance that it's a fair approximation of the Shannon entropy per symbol of the source. Knowing the data set length and the actual number of possible symbols (which depends heavily on how the input symbols are gathered, and what their physical source is) would help build an informed opinion on that.