Working out the following definition of the Cyclotomic Polynomial $$ {\displaystyle \Phi _{n}(x)=\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(x-e^{2i\pi {\frac {k}{n}}}\right),} $$ you'll get $$ {\displaystyle x^{\varphi(n)}-x^{\varphi(n)-1}\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} + \dots + 1 $$
You'll spot right away that the next to leading coefficient gives the Möbius function: $$ {\displaystyle \mu (n)=\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} $$
For the final $1$, you'll get: $$ \prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(-e^{2i\pi {\frac {k}{n}}}\right)=\left(e^{\frac{2i\pi}{n} \sum_{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}k}\right )\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}(-1)=\left(e^{\frac{2i\pi}{n}\frac{n}{2}\varphi(n) }\right )(-1)^{\varphi(n)}=+1, $$ which is expected since the coefficients are symmetric and $x^{\varphi(n)}$ also has $+1$.
Do the other terms also have number theoretical interpretation?
UPDATE:
A first partial result found at OEIS: Coefficient of $x^2$ in the $n$-th cyclotomic polynomial.
If $n$ is odd, $a(n) = 1/2 (\mu(n)^2-\mu(n))$, if $n$ is even, $a(n) = 1/2 (\mu(n)^2-\mu(n)) - \mu(n/2)$, where $\mu$ is Möbius function...
As Chris pointed out in his answer, the coefficient of the $m$th term of a cyclotomic polynomial is the elementary symmetric polynomial $e_m$ in the primitive $n$-th roots of unity, $\zeta_1,\zeta_2,...,\zeta_{\varphi(n)}$.
And $e_m$ can be written in the following way:
$${\displaystyle { \begin{aligned} e_{m}&={\frac {(-1)^{m}}{m!}}B_{m}(-p_{1},-1!p_{2},-2!p_{3},\ldots ,-(m-1)!p_{m})\\ &=(-1)^{m}\sum _{r_{1}+2r_{2}+\cdots +mr_{n}=m \atop r_{1}\geq 0,\ldots ,r_{m}\geq 0}\prod _{i=1}^{m}{\frac {(-p_{i})^{r_{i}}}{r_{i}!i^{r_{i}}}}\\ \end{aligned}}} $$
where the $B_n$ is the complete exponential Bell polynomial.
Now $p_k(\zeta_1,\zeta_2,...\zeta_{\varphi(n)}) $ can be seen as the Ramanujan sum evaluated at $k$: $$ p_k(\zeta_1,\zeta_2,...\zeta_{\varphi(n)}) =\sum _{\stackrel {1\leq t\leq n}{\gcd(t,\,n)=1}}e^{2\pi i{\frac {t}{n}k}}=c_n(k) $$
So $$ {\displaystyle e_{m}={\frac {(-1)^{m}}{m!}}B_{m}(-c_n(1),-1!c_n(2),-2!c_n(3),\ldots ,-(m-1)!c_n(m))} $$
and e.g. $$e_2=(-1)^2\left( \frac{(-c_1)^2}{2!1^2} + \frac{(-c_2)^1}{1!2^1} \right) = \frac12 \left( c_1^2-c_2 \right) \\ \color{grey}{\; \; \overset{?}{=} \frac12 \Big(\mu(n)^2-\underbrace{(\mu(n)+2(1-\bmod(n,2))\mu(n/2))}_{\overset{?}{=}c_2}\Big) }$$