Are $n$ vectors are orthogonal if performing the inner product of all $n$ vectors at once yields zero? Are $n$ vectors are orthogonal if performing the inner product of all $n$ vectors at once yields zero?
In other words, could I say that $\hat{i} \perp \hat{j} \perp \hat{k}$?
For example, suppose I have the three vectors $\hat{i} = <2, 0, 4>; \hat{j}= <0, 1, 0>; \hat{k} = <2, 0, 1>$. Their inner product would be $2(0)(2) + 0(1)(0) + 4(0)(1) = 0$. Can I say that the three vectors $\hat{i}, \hat{j}, $ and $ \hat{k}$ are orthogonal, or do I have to multiply out each pair?
 A: No, you cannot assume that. Inner products (i.e., dot products) can only be conducted with two vectors at a time. Hope this helps!
A: As already mentioned that the inner product is a bilinear form, and therefore the “inner product” of $n$ vectors is not defined. But, I find it interesting to ask what if we define “fake inner product” as you have done, by taking coordinate wise multiplication and summing; can we then conclude that the fake inner product means all $n$ vectors are mutually orthogonal? 
Well, in General it is still false. For example, take $(1,0), (0,1), (1,1)$ in $\mathbb{R}^2$. Note that the “fake inner product” is zero but clearly $(1, 1)$ and $(1,0)$ are not orthogonal in the usual sense.
One might object to it by observing that we are taking three vectors in $\mathbb{R}^2$. What if we have exactly $n$ vectors in $\mathbb{R}^n$ such that the “fake inner product” is zero. But, your claim is still false. Just take $e_i$ to be the standard basis vectors for $1\le i \le n-1$, and take any vector $v$ as the $n$th vector. It is clear that the fake inner product is 0, but we can choose $v$ which is not orthogonal to any of $e_i$. 
