As Robert Israel has pointed out, the sum of squares of $n$ independent random variables with a standard normal distribution has a chi-square distribution with $n$ degrees of freedom.
Take them from a normal distribution whose expectation is $\mu$ and whose standard deviation is $\sigma$, you have have
$$
\left(\frac{X_1-\mu}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-\mu}{\sigma}\right)^2
$$
has chi-square distribution with $n$ degrees of freedom.
So why might it appear that one of them is not counted? The answer to that comes from such results as this: Suppose instead of the population mean $\mu$, you subtract the sample mean $\overline X$. Then you have
$$
\left(\frac{X_1-\overline X}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-\overline X}{\sigma}\right)^2,\tag{1}
$$
and this has a chi-square distribution with $n-1$ degrees of freedom. In particular, if $n=1$, then the sample mean is just the same as $X_1$, so the numerator in the first term is $X_1-X_1$, and the sum is necessarily $0$, so you have a chi-square distribution with $0$ degrees of freedom.
Notice that in $(1)$, you have $n$ terms in the sum, not $n-1$, and they're not independent (since if you take away the exponents, you get $n$ terms that necessarily always add up to $0$) and the standard deviation of the fraction that gets squared is not actually $1$, but less than $1$. So why does it have the same probability distribution as if there were $n-1$ of them, and they were indepedent, and those standard deviations were each $1$? The simplest way to answer that may be this:
$$
\begin{bmatrix} X_1 \\ \vdots \\ X_n \end{bmatrix} = \begin{bmatrix} \overline X \\ \vdots \\ \overline X \end{bmatrix} + \begin{bmatrix} X_1 - \overline X \\ \vdots \\ X_n - \overline X \end{bmatrix}
$$
This is the decomposition of a vector into two components orthogonal to each other: one in a $1$-dimensional space and the other in an $n-1$ dimensional space. Now think about the spherical symmetry of the joint probability distribution, and about the fact that the second projection maps the expected value of the random vector to $0$.
Later edit:
Sometimes it might seem as if two of them are not counted. Suppose $X_i$ is a normally distributed random variable with expected value $\alpha+\beta w_i$ and variance $\sigma^2$, and they're independent, for $i=1,\ldots,n$. When $w_i$ is observable and $\alpha$, $\beta$, are not, one may use least-squares estimates $\hat\alpha$, $\hat\beta$. Then
$$
\left(\frac{X_1-(\alpha+\beta w_1)}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-(\alpha+\beta w_n)}{\sigma}\right)^2 \sim \chi^2_n
$$
but
$$
\left(\frac{X_1-(\hat\alpha+\hat\beta w_1)}{\sigma}\right)^2 + \cdots + \left(\frac{X_n-(\hat\alpha+\hat\beta w_n)}{\sigma}\right)^2 \sim \chi^2_{n-2}.
$$
A similar sort of argument involving orthogonal projections explains this.
One needs these results in order to derive things like confidence intervals for $\mu$, $\alpha$, and $\beta$.
The $x\ge0$ PDF $\tfrac{1}{2^{k/2-1}\Gamma(k/2)}x^{k-1}e^{-x^2/2}$ is asymptotic to $e^{-S}$ with$$S:=x^2/2-(k-1)\ln x+(k/2-1)\ln2+\ln\Gamma(k/2).$$As per Laplace's method, for large $k$ we can approximate the log-PDF as a quadratic, based on its first two derivatives. Since $S_x=x-\tfrac{k-1}{x}$ is $0$ at $x=\sqrt{k-1}$, and $S_{xx}=1+\tfrac{k-1}{x^2}>0$, this value of $x$ minimizes $S$ and maximizes the PDF. At this value of $x$,$$S=(k-1)/2-\tfrac12(k-1)\ln(k-1)+(k/2-1)\ln2+\ln\Gamma(k/2)$$and $S_{xx}=2$, so in general$$S\approx (k-1)/2-\tfrac12(k-1)\ln(k-1)+(k/2-1)\ln2+\ln\Gamma(k/2)+\left(x-\sqrt{k-1}\right)^2,$$and we've approximated the $\chi_k$ distribution as $N(\sqrt{k-1},\,\tfrac12)$. This Gaussian approximation can be restated as $\frac{\chi_k-\mu_k}{\sigma_k}$ being approximately Gaussian, with mean $0$ and variance $1$, and hence $\approx N(0,\,1)$.
Best Answer
A result due to Fisher states that the distribution of $\sqrt{2S_n}$ is approximately normal with mean $\sqrt{2n-1}$ and unit variance. Hence $\sqrt{2S_n}=\sqrt{2n-1}+Z_n$ where the distribution of $Z_n$ is approximately standard normal. Let $G$ denote the standard normal CDF. This means that $$ [S_n\leqslant x]=[\sqrt{2S_n}\leqslant \sqrt{2x}]=[Z_n\leqslant \sqrt{2x}-\sqrt{2n-1}], $$ hence $$ \mathrm P(S_n\leqslant x)\approx G(\sqrt{2x}-\sqrt{2n-1}). $$