Looking for ways how to calculate $\Phi_n(i)$ $Φ_n(1)$ and $Φ_n(−1)$ for the cyclotomic polynomials are well-known.
I am now looking for 
$$Φ_n(i)$$
 and/or 
$$Φ_n(−i)$$
with i the complex unit.
At this moments my endeavours result in intricate categorizing of values 
for $n$ e.g. as $4$, $2^k$, $4p$, $p^k$ with $p$ prime and considerations about
the prime factors of an odd $n$ being e.g. $\equiv1\mod4$ or/and $\equiv3\mod4$.
 A: The question has already been answered by Vladimir Dotsenko, but as Wolfgang Tintemann is still interested in a solution using Möbius inversion, let me expand the comments I made above.
Since $\Phi_n(-i)=\overline{\Phi_n(i)}$, it is enough to deal with $i$. We are going to prove
$$\Phi_n(i)=\begin{cases}
i-1&n=1,\\
i+1&n=2,\\
0&n=4,\\
p&n=4p^k,\\
i&n=p^k,\ p\equiv3\pod4,\text{ $k$ odd, or}\\
&n=2p^k,\ p\equiv3\pod 4,\text{ $k$ even,}\\
-i&n=p^k,\ p\equiv3\pod4,\text{ $k$ even, or}\\
&n=2p^k,\ p\equiv3\pod 4,\text{ $k$ odd,}\\
-1&n=p^kq^l\text{ or }n=2p^kq^l,\ p,q\equiv3\pod4,\\
1&\text{if $n$ is not of any of the forms above,}
\end{cases}$$
where $p,q$ are prime, and $k,l\ge1$. The first three cases can be checked immediately, hence let me assume that $n\ne1,2,4$ below.
By Möbius inversion of the formula $x^n-1=\prod_{d|n}\Phi_d(x)$, we have
$$\Phi_n(x)=\prod_{d|n}(x^d-1)^{\mu(n/d)}.$$
We cannot directly plug in $x=i$ as we might get $0/0$, but we can compute it as $\lim_{\epsilon\to0}\Phi_n(i+\epsilon)$. 
Since $(i+\epsilon)^d=i^d+O(\epsilon)$ for $4\nmid d$, and $(i+\epsilon)^d=1-id\epsilon+O(\epsilon^2)$ for $4\mid d$, we see that $\Phi_n(i+\epsilon)$ is
$$(1+O(\epsilon))
\prod_{\substack{d|n\\d\equiv1\pod4}}(i-1)^{\mu(\frac nd)}
   \prod_{\substack{d|n\\d\equiv3\pod4}}(-i-1)^{\mu(\frac nd)}
   \prod_{\substack{d|n\\d\equiv2\pod4}}(-2)^{\mu(\frac nd)}
   \prod_{\substack{d|n\\d\equiv0\pod4}}(-id\epsilon)^{\mu(\frac nd)}.$$
We can express $\pm i-1$ in terms of $\zeta_8=(1+i)/\sqrt2$ and reshuffle the terms to get
$$\tag{$*$}(1+O(\epsilon))
  (-\sqrt2)^{\sum_{d|n}\mu(\frac nd)}
  (\sqrt2)^{\sum_{2|d|n}\mu(\frac nd)}
  (2i\epsilon)^{\sum_{4|d|n}\mu(\frac nd)}
  \zeta_8^{s(n)}
  \prod_{4|d|n}(d/4)^{\mu(\frac nd)},$$
where
$$s(n)=\sum_{\substack{d|n\\d\equiv3\pod4}}\mu(n/d)
      -\sum_{\substack{d|n\\d\equiv1\pod4}}\mu(n/d).$$
Since $n\ne1$, $\sum_{d|n}\mu(n/d)=0$. Moreover, $\sum_{2|d|n}\mu(n/d)=\sum_{d\mid\frac n2}\mu(\frac n2/d)=0$ for $n$ even, and trivially $\sum_{2|d|n}\mu(n/d)=0$ for $n$ odd; likewise for $4$. Thus the first four terms of $(*)$ disappear as $\epsilon\to0$. If $4\nmid n$, the last term is vacuously $1$. On the other hand, if $4\mid n$, we have $s(n)=0$: all the $\mu(n/d)$ summands vanish on account of $n/d$ being divisible by $4$. Thus we arrive at
$$\tag{$*{*}$}
\Phi_n(i)=\begin{cases}
\zeta_8^{s(n)}&4\nmid n,\\
\prod_{d|m}d^{\mu(m/d)}&n=4m.
\end{cases}$$
Assume first $4\mid n$. We can simplify $(*{*})$ further to
$$\Phi_n(i)=m^{\sum_{d|m}\mu(m/d)}\prod_{d|m}(m/d)^{-\mu(m/d)}
=\prod_{d|m}d^{-\mu(d)}.$$
If $m=p^k$ is a prime power, this gives immediately $\Phi_n(i)=p$. On the other hand, if $m=p_1^{k_1}\cdots p_r^{k_r}$ with $r\ge2$, we have
$$\Phi_n(i)=\prod_{j=1}^rp_j^{\sum_{d|m_j}\mu(d)}=1,$$
where $m_j=m/p_j^{k_j}$.
Now assume $4\nmid n$, we need to compute $s(n)$ modulo $8$. Observe that if $n$ is odd, then $s(2n)=-s(n)$, hence $\Phi_{2n}(i)=\Phi_n(i)^{-1}$. It thus suffices to consider odd $n$.
If $n$ has a prime divisor $p\equiv1\pod4$, then each $d\mid n$ with $d\equiv3\pod4$ is paired with another one with opposite $\mu(n/d)$ by multiplying or dividing $d$ by $p$; likewise for $d\equiv1\pod4$. Thus in this case $s(n)=0$ and $\Phi_n(i)=1$.
Finally, assume that $n=p_1^{k_1}\cdots p_r^{k_r}$, where each $p_j\equiv3\pod4$. Let $t=\pm1$ be such that $n\equiv-t\pod4$. Then for every $I\subseteq\{1,\dots,r\}$, $d=n\prod_{i\in I}p_i^{-1}$ contributes $t(-1)^{|I|}\mu(n/d)=t$ to $s(n)$, hence $s(n)=t2^r$. Thus, $\Phi_n(i)=1$ if $r\ge3$, $\Phi_n(i)=-1$ if $r=2$, and $\Phi_n(i)=i^t=(-1)^{k_1+1}i$ if $r=1$.
A: Let me try to summarize what I said in the comment to your question and what I implicitly used in what did not fit in the comment. As you will instantly see, the answer is lengthy because of high-school level exercises you have to do along the way, not because some important ideas were missing.
I suggest to use the following properties of cyclotomic polynomials: 


*

*for $p$ prime, $k>1$, we have $\Phi_{mp^k}(x)=\Phi_{mp}(x^{p^{k-1}})$; 

*for $p$ prime and $m$ coprime with $p$, we have $\Phi_{mp}(x)=\frac{\Phi_m(x^p)}{\Phi_m(x)}$;

*for odd $m>1$ we have $\Phi_{2m}(x)=\Phi_m(-x)$.


The first of these, interpreted as saying that a primitive $mp^k$-th root of unity raised to the power $p^{k-1}$ is a primitive $mp$-th root of unity, is easy to understand directly, as is the second one that effectively says that numbers that become primitive $m$-th roots of unity once raised to the power $p$ are either primitive $m$-th roots of unity themselves of primitive $mp$-th roots of unity. The third property says that negatives of $m$-th roots of unity are $2m$-th roots of unity. 
(Iterating the first property, we easily obtain the formula $\Phi_n(x)=\Phi_{r}(x^{n/r})$ where $r$ is the product of all distinct prime divisors of $n$, which I mentioned in my comment).
Now, let us iterate these. 
First let us suppose that $n$ is odd. 
Let $p$ be a prime divisor of $n$ that is congruent to $1$ modulo $4$, and let $n=p^km$ where $\gcd(m,p)=1$. 
A. For $k=1$ we have $\Phi_{n}(x)=\frac{\Phi_m(x^p)}{\Phi_m(x)}$, so $\Phi_n(i)=\Phi_n(-i)=1$. (Since under our assumption on $p$ we have $i^p=i$).
B. For $k>1$ we have $\Phi_n(x)=\Phi_{mp}(x^{p^{k-1}})$, so $\Phi_n(i)=\Phi_{mp}(i)=\Phi_n(-i)=\Phi_{mp}(-i)=1$.
Let $p$ be a prime divisor of $n$ that is congruent to $3$ modulo $4$, and let $n=p^km$ where $\gcd(m,p)=1$. 
C. For $k=1$ we have $\Phi_{n}(x)=\frac{\Phi_m(x^p)}{\Phi_m(x)}$, so $\Phi_n(i)=\frac{\Phi_m(-i)}{\Phi_m(i)}$ and $\Phi_n(-i)=\frac{\Phi_m(i)}{\Phi_m(-i)}$. (Since under our assumption on $p$ we have $i^p=i^3=-i$).
D. For $k>1$ we have $\Phi_n(x)=\Phi_{mp}(x^{p^{k-1}})$, so when $k$ is even we have $\Phi_n(i)=\Phi_{mp}(-i)=\frac{\Phi_m(i)}{\Phi_m(-i)}$ and $\Phi_n(-i)=\Phi_{mp}(i)=\frac{\Phi_m(-i)}{\Phi_m(i)}$, and when $k$ is odd we have $\Phi_n(i)=\Phi_{mp}(i)=\frac{\Phi_m(-i)}{\Phi_m(i)}$ and $\Phi_n(-i)=\Phi_{mp}(-i)=\frac{\Phi_m(i)}{\Phi_m(-i)}$.
Now let us handle the case of even $n$. Let $n=2^km$, where $m>1$ is odd. 
E. For $k=1$ we have $\Phi_{n}(x)=\Phi_m(-x)$, so $\Phi_n(i)=\Phi_m(-i)$ and $\Phi_n(-i)=\Phi_m(i)$.
F. For $k=2$ we have $\Phi_{n}(x)=\Phi_{2m}(x^2)=\Phi_m(-x^2)$, so $\Phi_{n}(i)=\Phi_{n}(-i)=\Phi_m(1)$.
G. For $k>2$ we have $\Phi_{n}(x)=\Phi_{2m}(x^{2^{k-1}})$, so $\Phi_{n}(i)=\Phi_{n}(-i)=\Phi_{2m}(1)$, which is equal to $1$ for $m>1$ and to $2$ for $m=1$.
Finally, if $n=2^k$, we see that $\Phi_n(x)=x^{2^{k-1}}+1$, so $\Phi_n(i)$ is equal to $i+1$ for $k=1$, to $0$ for $k=2$, and to $2$ for $k>2$. Similarly, $\Phi_n(-i)$ is equal to $-i+1$ for $k=1$, to $0$ for $k=2$, and to $2$ for $k>2$.
It remains to use simple induction to give the complete answer. 
If $n=1$, then $\Phi_n(i)=i-1$ and $\Phi_n(-i)=-i-1$.
If $n$ is odd and has a prime divisor congruent to $1$ modulo $4$, then $\Phi_n(i)=\Phi_n(-i)=1$, this is established above.
If $n$ is odd and has just one prime divisor which is, in addition, congruent to $3$ modulo $4$, so that $n=p^k$, then $\Phi_n(i)=\frac{\Phi_1(i^{p^k})}{\Phi_1(i^{p^{k-1}})}$ which is equal to $\frac{\Phi_1(i)}{\Phi_1(-i)}=-i$ for even $k$ and to to $\frac{\Phi_1(-i)}{\Phi_1(i)}=i$ for odd $k$. Similarly, $\Phi_n(-i)$ is equal to $i$ for even $k$ and to $-i$ for odd $k$.
If $n$ is odd and has two distinct prime divisors $p$ and $q$, both congruent to $3$ modulo $4$, so that $n=p^aq^b$, then $\Phi_n(i)=\frac{\Phi_{q^b}(i^{p^a})}{\Phi_{q^b}(i^{p^{a-1}})}$, and by recalling from the previous case that $\Phi_{q^b}(i)=-\Phi_{q^b}(-i)$, we deduce that $\Phi_n(i)=-1$. Similarly, $\Phi_n(-i)=-1$.
If $n$ is odd and has at least three distinct prime divisors, all congruent to $3$ modulo $4$, so that $n=p^aq^bm$ with $m>1$, then $\Phi_n(i)=\frac{\Phi_{q^bm}(i^{p^a})}{\Phi_{q^bm}(i^{p^{a-1}})}$, and by induction and the previous case (saying that $\Phi_{p^aq^b}(i)=\Phi_{p^aq^b}(-i)$, we deduce that $\Phi_n(i)=1$. Similarly, $\Phi_n(-i)=1$.
If $n$ is even and is divisible by $2^k$ with $k\ge 3$, then $\Phi_n(i)=\Phi_n(-i)=1$, this is established above.
If $n$ is even, and $n=4m$ with odd $m>1$, then $\Phi_{n}(i)=\Phi_{n}(-i)=\Phi_m(1)$, as established above. The latter is equal to $1$ if $m$ is not a prime power, and is equal to $p$ if $m=p^k$, where $p$ is a prime. 
If $n$ is even, and $n=2m$ with odd $m>1$, then $\Phi_n(i)=\Phi_m(-i)$ and $\Phi_n(-i)=\Phi_m(i)$, and the values $\Phi_m(i)$, $\Phi_m(-i)$ are established above.
Finally, we already established that if $n=2^k$, then $\Phi_n(i)$ is equal to $i+1$ for $k=1$, to $0$ for $k=2$, and to $2$ for $k>2$. Similarly, $\Phi_n(-i)$ is equal to $-i+1$ for $k=1$, to $0$ for $k=2$, and to $2$ for $k>2$.
