Uncertainty in Uncertainty?
The objects on the l.h.s. of the position-momentum uncertainty relation $$ \Delta x \Delta p \geq \frac{\hbar}{2}$$ are standard deviations of quantum mechanical operators, defined for any operator $A$ by $$ \Delta A:=\sigma_A=\sqrt{\langle A^2 \rangle - \langle A\rangle^2}$$ where $\langle \dot{}\rangle$ denotes taking the expectation value with respect to a fixed state. Also, the quantum mechanical uncertainty relation has a priori nothing to do with imprecision of measurements, in particular, it does not refer to any limitation of the measurement apparatus, see e.g. this question. It refers to an intrinsic property of quantum states not having simultaneously well-defined classical values for all possible observables.
$\hbar$ is a constant, not a quantum mechanical operator, so this uncertainty relation means nothing for it. It just doesn't apply. All "uncertainty" on $\hbar$ is of the ordinary experimental kind, which has never stopped us from using fixed "true" values for constants in our theoretical considerations. Of course, for any computations that are sensitive to very small changes in $\hbar$ you'll need to take that experimental uncertainty into account, but that has nothing to do with quantum mechanics or the uncertainty principle, it's just how using experimentally measured values always works.
In your post you've intertwined two distinct notions:
"The Heisenberg Uncertainty Principle states that . . . there is always some uncertainty while measuring things."
The uncertainty while measuring things is called measurement error and it is due to the experimental apparatus.
The uncertainty described by Heisenberg says that there is a fundamental limit, expressed in terms of $\hbar$, to the precision with which certain pairs of physical properties of a particle, known as complementary variables, such as position $x$ and momentum $p$, can be known simultaneously.
Now, when we are attempting to measure the Planck's Constant, wouldn't there be some uncertainty? This should mean that I can never be able to measure the Planck's Constant to full accuracy, ever.
There will be uncertainty related to the experimental technique used. However, the uncertainty in the knowledge of ℏ does not change the Uncertainty principle and its consequences.
But if the amount of uncertainty depends on ℏ (which is uncertain), doesn't this lead to uncertainty of uncertainty?
The amount of uncertainty of the measurement does not depend on ℏ's uncertainty, quite the opposite the precision of the measurement determines the significant digits of the result.
If two complementary physical properties can describe perfectly the state of a quantum object. Then, to me $\frac{\hbar}{2}$ is some sort of nature's machine epsilon that sets a limit to the amount of information we may know about the state of the quantum object.
First of all, the uncertainity principle can be rigorously derived from the framework of quantum mechanics as given by Von Neumann's postulates. This uncertainity is different from the measurement error caused by experimental apparatus. The error due to experimental apparatus can be curtailed by performing different experimental procedures instead of a single one. e.g you can use the spectra of blackbody radiation, photoelectric effect, the Stern-Gerlach experiment and many others in order to get a bound on the experimental error in the determination of Planck constant. But the Heisenberg uncertainity is a characteristic of nature, and cannot be removed away by experimental accuracy.
In general, all physical constants/quantities are measured experimentally upto some error. That is a characteristic of not just the Planck constant but of all physically measureable quantities. That doesn't mean that the theory is invalidated.