[Edit 2014-08-13: Thanks to a comment by Peter Shor, I have changed my estimate of the asymptotic growth rate of this series.]
My belief is that $\lim_{n\to\infty} \sum_{i<n} \Pr(E_i)$ grows as $\sqrt{n}$. I do not have a proof but I think I have a convincing argument.
Let $B_i = f(i)$ be a random variable that gives the number of balls in bin $i$. Let $B_{i,j} = \sum_{k=i}^j B_k$ be a random variable that gives the total number of balls in bins $i$ through $j$ inclusive.
You can now write $\Pr(E_i) = \sum_{b<j} \Pr(E_j \wedge B_{1,j} = b) \Pr(E_i \mid E_j \wedge B_{1,j} = b)$ for any $j < i$. To that end, let's introduce the functions $\pi$ and $g_i$.
$$\pi(j, k, b) = \Pr(B_j = k \mid B_{1,j-1} = b) = \binom{n-b}{k}\left(\frac{1}{n-j+1}\right)^k\left(\frac{n-j}{n-j+1}\right)^{n-b-k}$$
$$\begin{aligned}
g_i(j, k, b) \; &= \Pr(E_i \wedge B_{j,i} \le k \mid E_{j-1} \wedge B_{1,j-1} = b) \\
&= \begin{cases}
0 & k < 0 \\
1 & k >= 0 \wedge j > i \\
\sum_{l=0}^{j-b-1} \pi(j, l, b) g_i(j + 1, k - l, b + l) & \mathrm{otherwise}
\end{cases}\end{aligned}$$
We can write $\Pr(E_i)$ in terms of $g_i$:
$$ \Pr(E_i) = g_i(1, i - 1, 0) $$
Now, it's clear from the definition of $g_i$ that
$$ \Pr(E_i) = \frac{(n-i)^{n-i+1}}{n^n}h_i(n) $$
where $h_i(n)$ is a polynomial in $n$ of degree $i - 1$. This makes some intuitive sense too; at least $n - i + 1$ balls will have to be put in one of the $(i+1)$th through $n$th bins (of which there are $n-i$).
Since we're only talking about $Pr(E_i)$ when $n\to\infty$, only the lead coefficient of $h_i(n)$ is relevant; let's call this coefficient $a_i$. Then
$$ \lim_{n\to\infty} \Pr(E_i) = \frac{a_i}{e^i} $$
How do we compute $a_i$? Well, this is where I'll do a little handwaving. If you work out the first few $E_i$, you'll see that a pattern emerges in the computation of this coefficient. You can write it as
$$ a_i = \mu_i(1, i-1, 0) $$
where
$$ \mu_i(j, k, b) = \begin{cases}
0 & k < 0 \\
1 & k >= 0 \wedge i > j \\
\sum_{l = 0}^{j-b-1} \frac{1}{l!} \mu_i(j + 1, k - l, b+ l) & \mathrm{otherwise}
\end{cases}$$
Now, I wasn't able to derive a closed-form equivalent directly, but I computed the first 20 values of $Pr(E_i)$:
N a_i/e^i
1 0.367879
2 0.270671
3 0.224042
4 0.195367
5 0.175467
6 0.160623
7 0.149003
8 0.139587
9 0.131756
10 0.12511
11 0.119378
12 0.114368
13 0.10994
14 0.105989
15 0.102436
16 0.0992175
17 0.0962846
18 0.0935973
19 0.0911231
20 0.0888353
Now, it turns out that
$$ \DeclareMathOperator{\Pois}{Pois} \Pr(E_i) = \frac{i^i}{i! e^i} = \Pois(i; i) $$
where $\Pois(i; \lambda)$ is the probability that a random variable $X$ has value $i$ when it's drawn from a Poisson distribution with mean $\lambda$. Thus we can write our sum as
$$ \lim_{n\to\infty} \sum_{i=1}^n \Pr(E_i) = \sum_{x = 1}^{\infty} \frac{x^x}{x!e^x} $$
Wolfram Alpha tells me this series diverges. Peter Shor points out in a comment that Stirling's approximation allows us to estimate $\Pr(E_i)$:
$$ \lim_{n\to\infty} \Pr(E_x) = \frac{x^x}{x!e^x} \approx \frac{1}{\sqrt{2 \pi x}}$$
Let
$$ \phi(x) = \frac{1}{\sqrt{2 \pi x}} $$
Since
- $\lim_{x\to\infty}\frac{\phi(x)}{\phi(x+1)} = 1$
- $\phi(x)$ is decreasing
- $\int_1^n \phi(x)dx \to \infty$ as $n \to \infty$
our series grows as $\int_1^n \phi(x) dx$ (See e.g. Theorem 2). That is,
$$\sum_{i=1}^n Pr(E_i) = \Theta\left(\sqrt{n}\right)$$