$x \odot y \odot z$ is the negation of $(x\odot y) \odot z$? Since 'XNOR: $\odot$' is a $2$-input operator, normally, I think $x \odot y \odot z$ means either $(x\odot y) \odot z$ or $x\odot (y \odot z)$, which are equivalent, why Wolfram Alpha interprets $x \odot y \odot z$ as $3$-input XNOR $?$ Which is $(x \oplus y)\odot z$:

Now we have: (they are negation of each other)
$$x \odot y \odot z\ne(x\odot y) \odot z$$
Question:
(I was trying to answer this question, it also mentioned $(⊕)⊙≠⊙⊙$, which I would agree, but Wolfram Alpha says it's not.)
Is the notaion of $x \odot y \odot z$ commonly used for $3$-input XNOR that $(x \odot y)\oplus z$?
 A: Wolfram Alpha is apparently interpreting $x \text{ XNOR } y \text{ XNOR } z$ as $\text{NOT } (x \text{ XOR } y \text{ XOR } z)$
where $x \text{ XOR } y \text{ XOR } z$ is to be interpreted as $x \oplus y \oplus z$, which is equivalent to both $(x \oplus y) \oplus z$ and $x \oplus (y \oplus z)$: these expressions are all true if and only if an odd number of its terms are true.
Now, I would agree with you that it would make sense that $x \odot y \odot z$ should be equivalent to $(x\odot y) \odot z$ or $x\odot (y \odot z)$, just as $x \oplus y \oplus z$ is equivalent to both $(x \oplus y) \oplus z$ and $x \oplus (y \oplus z)$: these expressions are all true if and only if an even number of its terms are false.
Interpreting $x \odot y \odot z$ as such (i.e. interpreting $x \odot y \odot z$ as either $(x\odot y) \odot z$ or $x\odot (y \odot z)$), your truth-table shows that $x \odot y \odot z$ (second or third column) and $\neg (x \oplus y \oplus z)$ (first column) are not the same. In fact, they are the negation of each other, which shows that 
$$x \odot y \odot z \Leftrightarrow x \oplus y \oplus z$$
... which may be surprising, but note that this is the case in virtue of there being an odd number of terms involved, for we have that an odd number of its terms are true if and only if an even number of its terms are false.
Indeed, when we move to an even number of terms (say, four), we get:
$$x \odot y \odot z \odot w \Leftrightarrow \neg(x \oplus y \oplus z \oplus w)$$
Of course, WolframAlpha will interpret $x_1 \odot x_2 \odot \cdots \odot x_n$ for any $n$ (whether even or odd) as: $$x_1 \odot x_2 \odot \cdots \odot x_n \Leftrightarrow \neg(x_1 \oplus x_2 \oplus \cdots \oplus x_n)$$, and so this time it will agree with this.
In fact, one advantage of WolframAlpha's interpretation is that with that, you will never have that:
$$x_1 \odot x_2 \odot \cdots \odot x_n \Leftrightarrow x_1 \oplus x_2 \oplus \cdots \oplus x_n$$
Another argument one could make is that we already have an operator that works like the $\text{XNOR}$, namely the $\leftrightarrow$. And, the $\leftrightarrow$ does typically generalize such that:
$$x \leftrightarrow y \leftrightarrow z \Leftrightarrow (x\leftrightarrow y) \leftrightarrow z \Leftrightarrow  x\leftrightarrow (y \leftrightarrow z)$$
In other words, under the earlier interpretation, the $\odot$ is simply the same as the $\leftrightarrow$, and so one could argue that that gicves us room to reinterpret $\odot$ differently ... as WolframAlpha does.
However, there is also a pretty strong argument against doing what WolframAlpha does. The $\oplus$ and $\odot$ symbols are clearly based on the $+$ and $\odot$ symbols, i.e. the symbols for $\text{OR}$ and $\text{AND}$. Indeed, one can think of the $\odot$ as an 'exclusive $\text{AND}$ or '$\text{XAND}$': something that would be true if the terms on both sides agree in truth-value.
Moreover, the $\text{OR}$ and $\text{AND}$ are each others dual. That is, we have that $\neg (p \lor q) \Leftrightarrow \neg p \land \neg q$, and this generalizes to their 'multi'-versions:
$$\neg \bigwedge \{x_1, x_2, ... , x_n\} \Leftrightarrow \bigvee \{\neg x_1, \neg x_2, ... , \neg x_n\}$$
The same is true for the $\text{NOR}$ and $\text{NAND}$: they are also duals of each other (though since they are not associative, there is no generalized version).
So, it would make sense to have the $\text{XOR}$ and $\text{XAND}$ (i.e. $\oplus$ and $\odot$) be duals of each other too and, since they are associative, demand that:
$$\neg \bigodot \{x_1, x_2, ... , x_n\} \Leftrightarrow \bigoplus \{\neg x_1, \neg x_2, ... , \neg x_n\}$$
But that only holds under the first interpretation, and not (at least not for odd $n$) for WolframAlpha's interpretation.
In sum, I would say that Wolfram's interpretation is 'unusual' at best.
And so in answer to your question:

Is the notaion of $x \odot y \odot z$ commonly used for $3$-input XNOR that $(x \odot y)\oplus z$?

(which I think is better phrased as:)

Is the notation of $x \odot y \odot z$ commonly used for $3$-input XNOR that $\neg (x \oplus y \oplus z)$?

I would say probably not
