What you might want to use as your weight given your graph is using a three-part piecewise function: $$ z=\begin{cases}
0 & x<0\\
1-0.08x & 0\leq x\leq10 \\
0 & 10\leq x
\end{cases}
$$ where $x$ denotes days past and $z$ denotes the weight. The first part prevents returning a value to weigh anything that happens, the second makes the linear regression and the third creates the cutoff at the end of the graph. After finding the weight, you calculate the value of a certain point as you would for any arithmetic mean. For example, for $2$ days past the $200$ value, you would find $z$ for the $200$ point as $1-0.08\cdot2=0.84$, and the weights for the $1$s being $1,0.92,0.76,0.68$ e.t.c.
This fitting problem can be equivalently rewritten as fitting function of form:
$$ f(x) = K \sin(\omega x) + L \cos(\omega x) + C $$
And your original $A$ is just $A =\sqrt{K^2+L^2}$
This reduces it to just ordinary least squares problem. We get least squares estimators for $K,L$ from the equation
$$\begin{bmatrix}
K \\
L \\
C
\end{bmatrix} = (X^TX)^{-1}X^{T}y$$
Where $X$ is matrix formed by values of $\sin(\omega x),\cos(\omega x), 1$ evaluated at consecutive values of $x$ coordinate of your observations and $y$ are values of said observations.
This way you can see that $K,L$ are some linear functions of $y_i$. I'm not sure what Poissonian error bar is but in general finding variance of sum of variables can be done if we know variance of individual variables.
Assuming $y_i$ uncorrelated we get:
$$ {Var}(K) = (1,0,0) . (X^TX)^{-1}X^{T} . Var(y_i)$$
and analog for $L$.
And $C$ is even simpler as it is just the mean value of the observations.
This way we have found $Var(K),Var(L),Var(C)$ assuming these are small enough you can just propagate error in the formula
$$ V = \frac{\sqrt{K^2+L^2}}{C} $$
either by direct computation or using this helpful lookup.
Best Answer
You can fit $$ y = M(1- e^{-ax}) $$ where $M$ is the maximum at which the plateau ends up and $a$ governs how fast it gets there.
You can probably get Excel to find the best values of $M$ and $a$, or just play until the graph looks right.