How to analyze the off-by-one error's impact on the entropy of the result?
First of all the first Entropy calculation is not correct. Recall the definition of (Shannon) entropy:

$$H(X) = -\sum_{i=1}^n {\mathrm{P}(x_i) \log_{\,b} \mathrm{P}(x_i)}$$

If uniformly chosen a digits entropy is 

$$H(X) = -\frac{1}{10} \log_2\left(\frac{1}{10}\right) = 3.3219280...$$


Case $&lt;250$ with 16 digits;

\begin{align}
H_1(X) &amp;= 16 \cdot - \sum_{i=1}^{10}{\frac{1}{10} \log_2\left(\frac{1}{10}\right)}\\
H_1(X) &amp;= 16 \cdot 10 \cdot -\frac{1}{10} \log_2\left(\frac{1}{10}\right) = 53.15084951...\\
\end{align}

See the calculation at Wolfram Alpha 
Case $\leq250$ with 16 digits; 

$$H_2(X) = 16 \cdot - \left( \frac{26}{251} \cdot \log_2\left(\frac{26}{251}\right)+9 \cdot \frac{25}{251} \cdot 
\log_2\left(\frac{25}{251}\right) \right) = 53.14921795...$$

See the calculation at Wolfram Alpha


It seems this is a mistake in the book. The current errata doesn't include this.

The difference is tiny but can make a difference since 0 will occur more and that is distinguishable. For example, take a 1.000.000 sample occurrence of 0 is around  103,500 – approximately 10.36%.

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For the visualization of the difference read this nice blog Anatomy of a pseudorandom number generator – visualizing Cryptocat’s buggy PRNG