Higher order empirical entropy is not the entropy of the empirical distribution?
No, as your counterexample shows, it is not true.  I don't understand why you previously thought it was true, either, so it's hard to say what might have gone wrong with your previous intuition.

We can say that

$$H(X) = H(X_1) + H(X_2 | X_1) + H(X_3 | X_1,X_2) + H(X_4 | X_1,X_2,X_3),$$

where $X_i$ denotes the $i$th character in $X$.  that much is true.

In the example you gave, $H(X_1) = 0$, $H(X_2|X_1) = 1$, $H(X_3|X_1,X_2) = 0.5$, $H(X_4 | X_1,X_2,X_3) = 0.75$ (see footnote * for explanation).  Thus $H(X) = 1 + 0.5 + 0.75 = 2.25$.

Do notice that $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ both depend on $i$.  This might be important.

Also, notice that $H(X_{i+1}|X_1,\dots,X_i)$ is unrelated to $H_1(x)$.

Perhaps you were expecting $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ to be the same for all $i$?  If that were true, there might be a relationship between $H_1(x)$ and $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ (or something similar to them)... but it's not.



Footnote *: How did I calculate those?  I used the formula

$$H(X_{i+1}|X_1,\dots,X_i) = \sum_v \Pr[(X_1,\dots,X_i) = v] \times H(X_{i+1}|(X_1,\dots,X_i) = v).$$

For $i=1,2$, we only need to consider two values of $v$; for $i=3$, we only need to consider three values of $v$.