Let $\mathbb{T} = [- \pi , \pi ]$.
Problem: I am having considerable trouble intuitively understanding how it could ever be that
$$ \lim_{n \to \infty} \left| \hat{f}(n) \right| = 0 $$
since the only way this could converge to $0$ is if
$$ \lim_{n \to \infty} \left| {1 \over 2 \pi} \int_{- \pi}^{\pi} f(\theta) e^{-in \theta} d\theta \right| = 0 $$
and hence
$$ \lim_{n \to \infty} \left| {1 \over 2 \pi} \int_{- \pi}^{\pi} f(\theta) d\theta \right| = 0 $$
Here $\int_{- \pi}^{\pi} f( \theta ) d \theta$ is a constant that may or may not equal $0$. If it does equal $0$, then all of the $\hat{f}(n)$ are equal to $0$; otherwise, they are all non-equal to $0$. But I must be wrong somewhere here, because if it really boiled down to this then how could the $\hat{f}(n)$ ever be useful in a Fourier Series?
Specific Example: The constant function $f = 1$ is periodic and $C^k$ for all $k \in \mathbb{N}$. So its fourier series should converge uniformly to $1$ for all $\theta \in \mathbb{T}$ (this is a theorem in a textbook I'm reading on Harmonic Analysis).
Here, I fail to see how the $\hat{f}(n)$ can converge to $0$. This is because
$$ \left| \hat{f}(n) \right| = \left| {1 \over 2 \pi} \int_{- \pi}^{\pi} f(\theta) e^{-in \theta} d\theta \right| \le {1 \over 2 \pi} \int_{- \pi}^{\pi} \left| f(\theta) \right| d\theta = {1 \over 2 \pi} \int_{- \pi}^{\pi} \left| 1 \right| d\theta = 1 $$
Hence
$$ \lim_{n \rightarrow \infty} |\hat{f}(n)| = 1 $$
and hence for any $\theta \in \mathbb{T}$:
$$ \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{in\theta} = \sum_{n \in \mathbb{Z}} 1 e^{in\theta} \ne 1 $$
This is a contradiction, so clearly something is wrong with my reasoning. But what is it?
Unfortunately, you have a lot of false claims or misunderstandings here.
You say that $\int_{\mathbb{T}} f(\theta) e^{-in\theta} \to 0$ implies that $\int_{\mathbb{T}} f = 0$. This is incorrect, since the first will be true for any continuous function by the Riemann-Lebesgue lemma.
You say that if $\int_{\mathbb{T}} f = 0$ then $\hat{f}(n) = 0$ for all $n$. Again, this is not true: Take any $2\pi$-periodic function and shift it by a constant so that the function has average zero. It probably has non-zero Fourier coefficients at at least some index.
You conclude that for the constant function $1$, the Fourier coefficients tend to $1$ in modulus. However, the only conclusion that your work can show is that $\limsup_{n \to \infty} |\hat{f}(n)| \le 1$, which is true.
The key point here is that you're missing out on the cancellation provided by the oscillatory terms $e^{-in\theta}$. When you integrate these over $\mathbb{T}$, you get $0$ because of this. For a very handwavey argument for the Riemann-Lebesgue lemma, consider this:
If $f$ is constant, then the Fourier coefficients are zero for $n > 0$ because of this. If $f$ is uniformly continuous and $n$ is large, $f$ looks like a constant on a small interval. If $n$ is large enough, this interval contains a whole bunch of periods of $e^{-in \theta}$ and the cancellation leads to a small Fourier coefficient.