What is the minimal polynomial equation with integral coefficients that the area of the regular 11-gon with side lengths 1 satisfies?
For the area $A$ of regular $n$-gon with side-length $1$, we have
$$A = \frac{n}{4} \cot\frac{\pi}{n} \qquad\to\qquad \cos^2\frac{\pi}{n}= \frac{x}{x+n^2}$$
where $x := 16 A^2$. The Chebyshev polynomial $T_n(\cdot)$ is such that
$$T_n(\cos(\pi/n)) = \cos(\pi) = -1$$
Therefore, finding a (not-necessarily-minimal) polynomial satisfied by $x$ is a matter of eliminating $c := \cos(\pi/n)$ from the polynomial system
$$\begin{align}
(x+n^2)\;c^2 - x &= 0 \tag{1}\\[4pt]
T_n(c)+1 &= 0 \tag{2}
\end{align}$$
A computer algebra system can do this readily; Mathematica, for instance, has a Resultant[] command for this purpose. The general theory of resultants is a bit of computational overkill here, though. We can use $(1)$ to reduce $(2)$ to a linear equation in $c$, say $P c + Q = 0$; then we can write $c^2 = \frac{Q^2}{P^2}$, so that $P^2 x - Q^2 ( x + n^2 )$ gives the target polynomial. (Of course, unless/until a closed form is determined, the brute-force symbol manipulation remains something to be done with a CAS.)
These are the results of the brute-force operation ...
Factor[Resultant[ ChebyshevT[n,c] + 1, (x+n^2)c^2 - x, c ]]
... performed by Mathematica for various $n$ (with extraneous factors and powers left in for completeness): $$\begin{align} n = 3 &: \quad 81\;(x - 3)^2 \\[4pt] n = 4 &: \quad 4\;(x - 16)^4 \\[4pt] n = 5 &: \quad 625\;(x^2 - 50 x + 125)^2 \\[4pt] n = 6 &: \quad 4x^2\;(x-108)^4 \\[4pt] n = 7 &: \quad 2401\;(x^3 - 245 x^2 + 7203 x - 16807)^2 \\[4pt] n = 8 &: \quad 4\;(x^2 - 384 x + 4096)^4 \\[4pt] n = 9 &: \quad 6561(x-27)^2\;(x^3 - 729 x^2 + 72171 x -177147)^2 \\[4pt] n = 10 &: \quad 4 x^2\;(x^2 - 1000 x + 50000 )^4 \\[4pt] n = 11 &: \quad 14641\;\left(\; \begin{array}{c} x^5 - 1815 x^4 + 614922 x^3 \\ - 53146830 x^2 + 1071794405 x - 2357947691 \\ \end{array}\;\right)^2 \\[4pt] n = 12 &: \quad 4 (x - 144)^4\;(x^2 - 2016 x + 20736)^4 \\[4pt] n = 13 &: \quad 28561\;\left(\; \begin{array}{c} x^6 - 3718 x^5 + 2827539 x^4 - 637138788 x^3 \\ + 44865189655 x^2 - 827150951094 x + 1792160394037 \end{array} \;\right)^2 \end{align}$$
Edit. Since $n$ itself is part of the elimination process, it figures into the result(ant)s in ways that the above doesn't make clear. Here are the interesting factors, using explicit references to powers of $n$ among the coefficients.
$$\begin{align} n = 3 &: \quad x - n \\[4pt] n = 4 &: \quad x - n^2 \\[4pt] n = 5 &: \quad x^2 - 2 n^2 x + n^3 \\[4pt] n = 6 &: \quad x - 3 n^2 \\[4pt] n = 7 &: \quad x^3 - 5 n^2 x^2 - 3 n^4 x - n^5 \\[4pt] n = 8 &: \quad x^2 - 6 n^2 x + n^4 \\[4pt] n = 9 &: \quad x^3 - n^3 x^2 + 11 n^4 x - 3 n^5 \\[4pt] n = 10 &: \quad x^2 - n^3 x + 5 n^4 \\[4pt] n = 11 &: \quad x^5 - 15 n^2 x^4 + 42 n^4 x^3 - 30 n^6 x^2 + 5 n^8 x - n^9 \\[4pt] n = 12 &: \quad x^2 - 14 n^2 x + n^4 \\[4pt] n = 13 &: \quad x^6 - 22 n^2 x^5 + 99 n^4 x^4 - 132 n^6 x^3 + 55 n^8 x^2 - 6 n^{10} x + n^{11} \end{align}$$
It may be worth noting that $11$ is a common factor of most of the coefficients in the $n=13$ polynomial, and $3$ is a common factor in half of the coefficients in the $n=11$ polynomial.
It may also be worth noting that the degree of $n$ in the polynomials gives the sequence $$1, 2, 3, 2, 5, 4, 5, 4, 9, 4, 11, ...$$ which is an offset of OEIS's A216475: "The number of numbers coprime to and less than $n$, excluding 2." (The OEIS sequence begins at index $1$, so its reference to $n+2$ corresponds to our use of $n$ here.) The pattern continues through at least $n=25$ with minor fiddling in the cases of $n=15$, $21$, and $25$: in these cases, the power of $n$ is too big by $1$; we can "fix" that by writing, $n^{p+1}$ as $n\cdot n^p$ and treating the pulled-off $n$ as "just a number". (In the case of $n=25$, writing $n^{20}$ as $25n^{19}$ allows us to divide the polynomial by its leading $x$-coefficient, $5$, making it monic in $x$.)
By the way, the degree sequence for $x$ ... $$1, 1, 2, 1, 3, 2, 3, 2, 5, 2, 6, ...$$ ... appears as an offset of OEIS's A023022: the "half-totient" function, $\phi(n)/2$.
Edit 2. We can use this explicit formula for $T_n$ ... $$T_n(c) = c^n\;\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}(1-c^{-2})^k$$ ... to give formulas for the polynomials above (the ones with the extra factors and powers; determining the minimal polynomial from these is an exercise for the reader).
When $n=2m$, there's no appreciable work, since $c$ appears in even powers throughout, and $(2)$ simply becomes (after clearing denominators) $$\left(x+n^2\right)^m + \sum_{k=0}^{m} \binom{n}{2k}(-1)^k x^{m-k} n^{2k}$$
When $n=2m+1$, there's an extra power of $c$, so our manipulations involve squaring to get $$\left(x+n^2\right)^{n} - x\;\left(\;\sum_{k=0}^{m} \binom{n}{2k}(-1)^k x^{m-k} n^{2k} \; \right)^2 $$