My take on this notation agrees with David Speyer's comment. As Henning's answer notes, our answers are just different ways to look at the same thing. Often I resort to rules similar to his as the operative definition, but I find the my approach easier to motivate.
We proceed in two steps: (a.) understanding an expression involving asymptotic notation, and (b.) understanding the use of the equals sign in this notation.
Rule for Interpreting Expressions. Every (well-formed) expression involving standard functions and operations (e..g, ${ +, \times, -, \div, \exp, \log }$) and asymptotic notations $\{ O, o, \Omega, \omega, \Theta\}$ corresponds to a set of functions. [For simplicity, we will restrict ourselves to only the $O$ notation.] This set is built recursively exactly as one would expect:
$$
\begin{align*}
E_1 \circ E_2 &:= \{ f(n) \circ g(n) \quad : \quad f(n) \in E_1, g(n) = E_2 \}, \quad \circ \in \{ +, \times, -, \div \}; \\
\Psi (E) &:= \{ \Psi(f(n)) \quad : \quad f(n) \in E \}, \quad \Psi \in \{ \exp, \log \}. \\
\end{align*}
$$
This interpretation of expressions is perfectly natural; it's analogous to the Minkowski sum of two sets, extended for more general operations. We can even enrich this with a slightly more involved rule for $O$:
$$
O(E) := \bigcup_{f(n) \in E} O(f(n)).
$$
These rules are not complete; one needs to append the appropriate base cases when $E$, $E_1$ and $E_2$ are single functions rather than a set of functions. The most interesting base case is the expression $O(f(n))$ which is defined exactly as in the post. As a final point, we can agree to identity a single function $f(n)$ with the set $\{ f(n) \}$; this turns out to be very convenient in practice.
For example, an expression of the form $5 + O(n) + O(n^2)\log(O(n^3))$ stands for the set
$$
\{ 5 + f(n) + g(n) \log h(n) \ : \ f(n) \in O(n), \quad g(n) \in O(n^2), \quad h(n) \in O(n^3) \}.
$$
Similarly we can interpret the expression like $O(2^{n^2 + O(n)})$ by applying the $O(\cdot)$ rule twice.
Rules for Interpreting the Equals sign. When one asserts that $E_1 = E_2$ where $E_1$ and $E_2$ are sets represented by expressions like the above, one should always interpret it to mean that $E_1 \subseteq E_2$. E.g., $$5 + O(n) + O(n^2)\log(O(n^3)) \ \color{Red}{=} \ O(n^2\log n)$$ is really a shorthand for $$5 + O(n) + O(n^2)\log(O(n^3)) \ \color{Red}{\subseteq} \ O(n^2\log n) .$$
In addition, there is the case $f(n) = E$; e.g., $n = O(n^2)$. In this case, we can identity the function $f(n)$ with the set $\{ f(n) \}$ and apply the above interpretation. This tells us that $f(n) = E$ is the same as saying $f(n) \in E$ (because $f(n) \in E \iff \lbrace f(n) \rbrace \subseteq E$).
This is arguably unnatural since this clashes starkly with the standard interpretation of the equals sign as (set) equality. In fact, this is an inherent difficulty since we would expect any reasonable definition of equality to be symmetric:
$$
E_1 = E_2 \quad \stackrel{\color{Red}{(??)}}{\implies} \quad E_2 = E_1.
$$
However, as we all know, this is seldom true with the asymptotic notation. Thus the only way out seems to be to override the intuitive notion of equality in favor of a less natural one. [At this stage, it is worth pointing out that Henning's rules also resorts to redefining the equals sign suitably.]
A sample of properties. Many commonly used rules for manipulating asymptotic notation follow directly from the above interpretation. As a case study, I consider some of the properties already studied by Henning in his answer:
"Local scope" or Change of meaning. For example, In $O(n) + O(n) = O(n)$, you can substitute distinct functions $f(n)$, $g(n)$ and $h(n)$. The definition of "$+$" in the set-of-functions interpretation is clearly based on this idea.
The asymptotic notation is not symmetric. E.g., $O(n) = O(n^2)$ whereas $n^2 \stackrel{\color{Red}{?}}{=} O(n)$ is false. This is for the same reason that set inclusion is not.
However it is transitive; i.e., if $E_1 = E_2$ and $E_2 = E_3$, then $E_1 = E_3$. This simply follows from the transitivity of set inclusion.
Final words. One main complaint against the asymptotic notation seems to be that it is usually not symmetric as is implied by the equals sign. This is legitimate, but in this case, the real issue is with the abuse of equals sign, rather than the functions-as-sets idea. In fact, as far as I remember, I am yet to come across a single use of the asymptotic notation that is technically wrong even after one mentally replaces the equals sign with either $\in$ or $\subseteq$ as appropriate.
Just a hint as a possible way to go, through the Beta Function.
$$
\eqalign{
& f_{\,N,\,n} = \sum\limits_{k = 0}^{\min (N,n) - 1}
{{{k!\left( {n - 1 - k} \right)!\left( {N - 1 - k} \right)!} \over {\left( {N + n - 1 - k} \right)!}}} = \cr
& = n\sum\limits_{k = 0}^{\min (N,n) - 1}
{{{\left( {k + 1 - 1} \right)!\left( {n - k - 1} \right)!\left( {n - 1} \right)!\left( {N - k - 1} \right)!}
\over {\left( {n + 1 - 1} \right)!\left( {N + n - k - 1} \right)!}}} = \cr
& = n\sum\limits_{k = 0}^{\min (N,n) - 1}
{{\rm B}\left( {k + 1,n - k} \right){\rm B}\left( {n,N - k} \right)} \cr}
$$
Now, the integral representation of Beta looks to get clumsy.
On the other hand, Beta is rapidly decaying at growing of either of its arguments, so
it might be possible to try by truncating the sum.
Another way is that $F_{\, N}(x)$ actually reads
$$
\eqalign{
& F_{\,N} (x) = \sum\limits_{n = 0}^\infty {\sum\limits_{k = 0}^{\min (N,n) - 1}
{{{k!\left( {n - 1 - k} \right)!\left( {N - 1 - k} \right)!}
\over {\left( {N + n - 1 - k} \right)!}}} \;x^{\,n} } = \cr
& = \sum\limits_n {\sum\limits_k
{{{\left[ {0 \le k < n} \right]\left[ {0 \le k < N} \right]k!\left( {n - 1 - k} \right)!
\left( {N - 1 - k} \right)!} \over {\left( {N + n - 1 - k} \right)!}}x^{\,n} } } = \cr
& = \sum\limits_k {\left[ {0 \le k < N} \right]k!\left( {N - 1 - k} \right)!
\sum\limits_n {{{\left[ {k < n} \right]\left( {n - 1 - k} \right)!}
\over {\left( {N + n - 1 - k} \right)!}}x^{\,n} } } = \cr
& = \sum\limits_{0 \le k < N} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k < n} {{{1^{\,\overline {\,n - 1 - k\,} } }
\over {1^{\,\overline {\,N + n - 1 - k\,} } }}x^{\,n} } } = \cr
& = \sum\limits_{k = 0}^{N - 1} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k + 1 \le n} {{{1^{\,\overline {\,n - 1 - k\,} } }
\over {1^{\,\overline {\,n - 1 - k\,} } \left( {n - k} \right)^{\,\overline {\,N\,} } }}x^{\,n} } } = \cr
& = \sum\limits_{k = 0}^{N - 1} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k + 1 \le n} {{{x^{\,n} } \over {\left( {n - k} \right)^{\,\overline {\,N\,} } }}} } = \cr
& = \left( {N - 1} \right)!\left( {\sum\limits_{k = 0}^{N - 1} {{{x^{\,k} }
\over {\left( \matrix{ N - 1 \cr k \cr} \right)}}} } \right)
\left( {\sum\limits_{1 \le j} {{{x^{\,j} } \over {j^{\,\overline {\,N\,} } }}} } \right) \cr}
$$
where:
I am sure you can continue by yourself.
Best Answer
All students have difficulties when they meet the $O$- and the $o$-notations for the first time. Whenever such an $O$ appears it refers to a pre-agreed limit process for the independent variable, in the case at stake to $n\to\infty$. The statement $p(n)=O(n^k)\ (n\to\infty)$ does not mean that the (usually "complicated") function $p(n)$ is equal to some other function $O(\cdot)$, evaluated at $n^k$. Instead it expresses the (claimed or proven) fact that the function $p(n)$ under study, after division by $n^k$, stays bounded when $n\to\infty$; which is the same thing as saying that $p(n)=b(n)\cdot n^k$ where now $b(n)$ is a bounded function. In your example, each $n^i/n^k$ is $\leq 1$, so in fact $|p(n)|\leq C\cdot n^k$ with $C:=\sum_{i=1}^d |a_i|$.