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 $k>2$.