The claim is true for $\lambda = 0$, when $f(x) = x^n + 1$,
and for $\lambda = 1$, when $f(x) = (x+1)^n$. We show that this
implies the claim for all intermediate $\lambda$, by proving that
the number of zeros on the unit circle is a non-decreasing
function of $\lambda$.

Let $\lambda = e^t$ with $-\infty < t < 0$, and $x = e^{iu}$ with
$u \in {\bf R} \bmod 2\pi {\bf Z}$. Then
$$
f(x) = \exp\bigl[t\bigl(\frac{n}2\bigr)^2 + i\bigl(\frac{n}2\bigr)u\bigr]
\cdot F_t(u),
$$
where
$$
F_t(u) = \sum_{k=0}^n {n \choose k}
\exp\bigl[-t\bigl(k-\frac{n}{2}\bigr)^2 + i\bigl(k-\frac{n}{2}\bigr)u\bigr]
$$
is a real-valued function (the $k$ and $n-k$ terms are complex conjugates,
a symmetry already noted in the comments) and
$F_t(u+2\pi) = (-1)^n F_t(u)$ for all $u,t$.
Now the key observation is that $F$ satisfies the *heat equation*
$$
\frac{\partial}{\partial t} F_t(u) =
\sum_{k=0}^n -\frac{(2k-n)^2}{4} {n \choose k}
\exp\bigl[-t\bigl(k-\frac{n}{2}\bigr)^2 + i\bigl(k-\frac{n}{2}\bigr)u\bigr]
= \frac{\partial^2}{\partial^2 \! u} F_t(u).
$$
Thus as $t$ increases the periodic function $F_t(u)$ can lose zeros
(as when two zeros merge and then split into the complex plane)
but never gain them. Since $F_t$ has $n$ zeros in ${\bf R} / 2\pi {\bf Z}$
for $t<0$ of sufficiently large $|t|$,
while $F_0$ still has an $n$-fold zero at $u = \pi$,
it follows that $F_t$ must still have $n$ zeros for all $t<0$ as well.

*Added later:* Much the same argument proves that for $\lambda>1$
the polynomial has all roots on the negative real axis. Here we write
$x = e^u$, $\lambda = e^{-t}$, and $f(x) = \exp[-t(n/2)^2+(n/2)u] G_t(u)$,
and again check that $G_t(u)$ satisfies the heat equation because it is
a linear combination of heat-equation solutions
$\exp(tm^2+mu)$ with $m = k-\frac{n}{2}$.
For large $\lambda$ there are $n$ real roots because $f$ changes sign
near the negative root of
$$
{n \choose k-1} \lambda^{(k-1)(n-k+1)} x^{k-1}
+ {n \choose k} \lambda^{k(n-k)} x^k
$$
for each $k=1,\ldots,n$. For $\lambda = 1$ there is still an $n$-fold root
at $x=-1$, and of course there are no nonnegative roots for any $\lambda$.
Hence $f$ has $n$ negative real roots for all $\lambda > 1$.