This is an interesting question, and actually quite subtle. It gets at the core of the definition of interactions, as well as the assumptions underlying dummy coding of categorical variables.
OP's question is that, when we are testing for interactions between the levels as given above, why do we (seemingly!) not bother to take into account interactions between one variable and the other variable's reference level?
The short answer is that we actually are taking these into account, as I'll explain here. First, to simplify subsequent notation, let $Y_{AE}$ denote the random variable $Y|\{X_1=A\} \cap \{X_2=E\}$, that is, $Y$ conditional on observing levels $A$ and $E$ in variables $X_1$ and $X_2$, respectively.
Now, recall that if we claim that there is no interaction between $X_1$ and $X_2$, this is equivalent to asserting that
$$
E[Y_{AE}]-E[Y_{BE}] = E[Y_{AF}]-E[Y_{BF}] = E[Y_{AG}]-E[Y_{BG}] \qquad (Eq. 1)
$$
that is, the expected difference between $Y|X_1=A$ and $Y|X_1=B$ is unrelated to the level of $X_2$, as long as $X_2$ is held constant. Similarly, no interaction means that
$$ E[Y_{AE}]-E[Y_{AF}] = \cdots = E[Y_{DE}]-E[Y_{DF}]. $$
Now, OP correctly states that given our choice of reference levels $A$ and $E$, the definition of $\beta_1$ is the expected difference between $Y$ when $X_1=B$ versus $X_1=A$, given that $X_2$ is held constant at $E$. In other words,
$$ \beta_1 = E[Y_{AE}] - E[Y_{BE}], $$
but if there is no interaction between $X_1$ and $X_2$, it follows immediately from equation 1 above that
$$
\beta_1 = E[Y_{AE}]-E[Y_{BE}] = E[Y_{AF}]-E[Y_{BF}] = E[Y_{AG}]-E[Y_{BG}]
$$
In other words, if there is no interaction, then $\beta_1$ is sufficient to explain the differences in $Y$ based on levels of $X_1$ when $X_2$ is held constant. This is why the test for an interaction involves leaving $\beta_1$ in the model and testing whether the interaction terms are zero.
The full answer to your question involves a simultaneous derivation for $\beta_1$ through $\beta_5$, but from this example you hopefully get the idea of what's going on.
Notice that, if $x_2=0$, then
$$y= \beta_0 + \beta_1 x_1 + u$$
and that, if $x_2=1$, then
$$y= (\beta_0+\beta_2) + (\beta_1+\beta_3) x_1 + u,$$
so $(\beta_1+\beta_3)$ is how much one would expect $y$ to change for a unitary change in $x_1$ if $x_2=1$ and $\beta_1$ is how much one would expect $y$ to change for a unitary change in $x_1$ if $x_2=0$.
That is, the expected change in $y$ depend on whether $x_2=0$ or $x_2=1$.
In that context, one way to think of $\beta_3$ is as an additional (positive or negative) expected change in $y$ when $x_2=1$ beyond the expected change $\beta_1$. It is possible even that $\beta_1$ and $\beta_3$ cancel out one another, in which case $y$ is constant w.r.t. $x_1$ when $x_2=1$.
Since you mention no other transformation besides standardization, both $\beta_1$ and $\beta_3$ must be thought of as the expected change in $y$ in case of observing an increase of one standard deviation in the independent quantitative variable in its original scale.
Best Answer
With treatment/dummy coding, the value of the individual coefficient for a predictor involved in higher-level interactions is the value when all of its interacting predictors are at their reference levels. So the $\beta_2$ coefficient for $Z$ in your setup is the value when all its interacting binary predictors $W_1$, $W_2$ and $W_3$ are at their reference levels (taken to be 0) and when the continuous predictor $X$ has a value of 0.
You can see this for the $Z_{\text{high}}$ versus $Z_{\text{low}}$ comparison by examining all of the terms involving $Z$, as the other terms cancel in that comparison:
$$ \beta_2Z + \beta_3 XZ + \beta_6 ZW_1 + \beta_7 XZW_1 + \beta_{10} ZW_2 + \beta_{11} XZW_2 + \beta_{14} ZW_3 + \beta_{15} XZW_3. $$
Your proposed formula for $Z_{\text{high}}$ versus $Z_{\text{low}}$ at $W_1 = 0$, $\beta_2 + \beta_3X$, thus only holds for the situation where $W_2$ and $W_3$ are also at zero. Otherwise, you need to include the further interaction terms involving $Z$ and non-zero values of $W_2$ and/or $W_3$.