You seem to be suffering the usual confusion about what "linear regression" means. Nonlinearity in the $x$ values does not mean you're doing nonlinear regression. If you're talking about fitting the sort of polynomial you're talking about above under the simplest conventional assumptions, that's *linear* regression, notwithstanding the fact that you're fitting a curve rather than a line.

You're also using the lower-case $n$ to denote two different things in the same problem! Confusing at best!

Think about this:
$$
Y = X\beta + \varepsilon
$$
where $Y$ is an $n\times1$ vector, $X$ is an $n\times k$ matrix with (typically) $k \ll n$, and $\varepsilon$ is an $n\times 1$ random vector of "errors" that is normally distributed with expected value the $n\times 1$ column vector of $0$s and variance $\sigma^2$ times the $n\times n$ identity matrix. You can observe $X$ and $Y$ and you want to estimate $\beta$, which is of course a $k\times 1$ fixed but unobservable vector.

In the case where $Y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + \cdots + \beta_{k-1} x_i^{k-1} + \varepsilon_i$, we have
$$
X= \begin{bmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{k-1} \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{k-1} \end{bmatrix}
$$
Then the least-squares estimate of $\beta$ is $\hat\beta = (X^TX)^{-1}X^T Y$. The vector of fitted values of $Y$ is $\hat Y = HY = X(X^TX)^{-1}X^TY$. The "hat matrix" $H$ is an $n\times n$ symmetrix idempotent matrix of rank $k \ll n$. It's the orthogonal projection onto the column space of $X$. The vector of residuals is $\hat\varepsilon = Y - \hat Y = (I - H)Y$. This matrix $I-H$ represents the complementary orthogonal projection.

Now remember: If $A$ is an $m\times n$ fixed matrix and $B$ is an $\ell\times n$ fixed matrix then $E(AY) = A E(Y)$, $\operatorname{var}(AY) = A(\operatorname{var}(Y))A^T$, and $\operatorname{cov}(AY,BY) = A(\operatorname{cov}(Y,Y)B^T$.

Consequently we have
$$
\hat\beta \sim N_k (\beta, \sigma^2 (X^TX)^{-1}),
$$
(a $k$-dimensional normal distribution)
$$
\hat\varepsilon \sim N_n(0, \sigma^2(I - H))
$$
(an $n$-dimensional normal distribution with a variance of small rank, $k$), and
$$
\operatorname{cov}(\hat\beta, \hat\varepsilon) = 0
$$
(a $k\times n$ matrix of zeros).

Therefore the distribution of the sum of squares of observed residuals (not unboservable errors) is given by
$$
\frac{\|\hat\varepsilon\|^2}{\sigma^2} = \frac{\|(I-H)Y\|^2}{\sigma^2} \sim \chi^2_{n-k},
$$
a chi-square distribution with $n-k$ degrees of freedom, and this random vector is actually independent of the least-squares estimates $\hat\beta$, which is normally distributed.

Now if you think about the definition of the Student's t-distribution, you can derive various confidence intervals.